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Physics textbook for undergraduate Addis Ababa University students.

Physics textbook for undergraduate Addis Ababa University students.

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    AAU Physics Textbook AAU Physics Textbook Document Transcript

    • Physics for Scientists and Engineers (with PhysicsNOW and InfoTrac) Raymond A. Serway - Emeritus, James Madison University John W. Jewett - California State Polytechnic University, Pomona ISBN 0534408427 1296 pages Case Bound 8 1/2 x 10 7/8 Thomson Brooks/Cole © 2004; 6th Edition This best-selling, calculus-based text is recognized for its carefully crafted, logical presentation of the basic concepts and principles of physics. PHYSICS FOR SCIENTISTS AND ENGINEERS, Sixth Edition, maintains the Serway traditions of concise writing for the students, carefully thought-out problem sets and worked examples, and evolving educational pedagogy. This edition introduces a new co-author, Dr. John Jewett, at Cal Poly – Pomona, known best for his teaching awards and his role in the recently published PRINCIPLES OF PHYSICS, Third Edition, also written with Ray Serway. Providing students with the tools they need to succeed in introductory physics, the Sixth Edition of this authoritative text features unparalleled media integration and a newly enhanced supplemental package for instructors and students! Features A GENERAL PROBLEM-SOLVING STRATEGY is outlined early in the text. This strategy provides a series of steps similar to those taken by professional physicists in solving problems. This problem solving strategy is integrated into the Coached Problems (within PhysicsNow) to reinforce this key skill. A large number of authoritative and highly realistic WORKED EXAMPLES promote interactivity and reinforce student understanding of problem-solving techniques. In many cases, these examples serve as models for solving end-of-chapter problems. The examples are set off from the text for ease of location and are given titles to describe their content. Many examples include specific references to the GENERAL PROBLEM-SOLVING STRATEGY to illustrate the underlying concepts and methodology used in arriving at a correct solution. This will help students understand the logic behind the solution and the advantage of using a particular approach to solve the problem. About one-third of the WORKED EXAMPLES include new WHAT IF? extensions. CONCEPTUAL EXAMPLES include detailed reasoning statements to help students learn how to think through physical situations. A concerted effort was made to place more emphasis on critical thinking and teaching physical concepts in this new edition. Both PROBLEM-SOLVING STRATEGIES and HINTS help students approach homework assignments with greater confidence. General strategies and suggestions are included for solving the types of problems featured in the worked examples, end-of-chapter problems, and PhysicsNow. This feature helps students identify the essential steps in solving problems and increases their skills as problem solvers. END-OF-CHAPTER PROBLEMS – An extensive set of problems is included at the end of each chapter. Answers to odd-numbered problems are given at the end of the book. For the convenience of both the student and instructor, about two thirds of the problems are keyed to specific sections of the chapter. All problems have been carefully worded and have been checked for clarity and accuracy. Solutions to approximately 20 percent of the end-of-chapter problems are included in the Student Solutions Manual and Study Guide. These problems are identified with a box around the problem number. Serway and Jewett have a clear, relaxed writing style in which they carefully define new terms and avoid jargon whenever possible. The presentation is accurate and precise. The International System of units (SI) is used throughout the book. The U.S. customary system of units is used only to a limited extent in the problem sets of the early chapters on mechanics.
    • Table of Contents Part I: MECHANICS 1. Physics and Measurement. Standards of Length, Mass, and Time. Matter and Model Building. Density and Atomic Mass. Dimensional Analysis. Conversion of Units. Estimates and Order-of-Magnitude Calculations. Significant Figures. 1 2 2. Motion in One Dimension Position, Velocity, and Speed. Instantaneous Velocity and Speed. Acceleration. Motion Diagrams. One-Dimensional Motion with Constant Acceleration. Freely Falling Objects. Kinematic Equations Derived from Calculus. General Problem-Solving Strategy. 23 3. Vectors. Coordinate Systems. Vector and Scalar Quantities. Some Properties of Vectors. Components of a Vector and Unit Vectors. 58 4. Motion in Two Dimensions The Position, Velocity, and Acceleration Vectors. Two-Dimensional Motion with Constant Acceleration. Projectile Motion. Uniform Circular Motion. Tangential and Radial Acceleration. Relative Velocity and Relative Acceleration. 77 5. The Laws of Motion The Concept of Force. Newton's First Law and Inertial Frames. Mass. Newton's Second Law. The Gravitational Force and Weight. Newton's Third Law. Some Applications of Newton's Laws. Forces of Friction. 111 6. Circular Motion and Other Applications of Newton's Laws Newton's Second Law Applied to Uniform Circular Motion. Nonuniform Circular Motion. Motion in Accelerated Frames. Motion in the Presence of Resistive Forces. Numerical Modeling in Particle Dynamics. 150
    • 7. Energy and Energy Transfer Systems and Environments. Work Done by a Constant Force. The Scalar Product of Two Vectors. Work Done by a Varying Force. Kinetic Energy and the Work--Kinetic Energy Theorem. The Non-Isolated System--Conservation of Energy. Situations Involving Kinetic Friction. Power. Energy and the Automobile. 181 8. Potential Energy Potential Energy of a System. The Isolated System--Conservation of Mechanical Energy. Conservative and Nonconservative Forces. Changes in Mechanical Energy for Nonconservative Forces. Relationship Between Conservative Forces and Potential Energy. Energy Diagrams and Equilibrium of a System. 217 9. Linear Momentum and Collisions Linear Momentum and Its Conservation. Impulse and Momentum. Collisions in One Dimension. Two-Dimensional Collisions. The Center of Mass. Motion of a System of Particles. Rocket Propulsion. 251 10. Rotation of a Rigid Object about a Fixed Axis Angular Position, Velocity, and Acceleration. Rotational Kinematics: Rotational Motion with Constant Angular Acceleration. Angular and Linear Quantities. Rotational Kinetic Energy. Calculation of Moments of Inertia. Torque. Relationship Between Torque and Angular Acceleration. Work, Power, and Energy in Rotational Motion. Rolling Motion of a Rigid Object. 292 11. Angular Momentum The Vector Product and Torque. Angular Momentum. Angular Momentum of a Rotating Rigid Object. Conservation of Angular Momentum. The Motion of Gyroscopes and Tops. Angular Momentum as a Fundamental Quantity. 336 12. Static Equilibrium and Elasticity The Conditions for Equilibrium. More on the Center of Gravity. Examples of Rigid Objects in Static Equilibrium. Elastic Properties of Solids. 362
    • 13. Universal Gravitation Newton's Law of Universal Gravitation. Measuring the Gravitational Constant. Free-Fall Acceleration and the Gravitational Force. Kepler's Laws and the Motion of Planets. The Gravitational Field. Gravitational Potential Energy. Energy Considerations in Planetary and Satellite Motion. 389 14. Fluid Mechanics Pressure. Variation of Pressure with Depth. Pressure Measurements. Buoyant Forces and Archimedes's Principle. Fluid Dynamics. Bernoulli's Equation. Other Applications of Fluid Dynamics. 420 Part II: OSCILLATIONS AND MECHANICAL WAVES 451 15. Oscillatory Motion Motion of an Object Attached to a Spring. Mathematical Representation of Simple Harmonic Motion. Energy of the Simple Harmonic Oscillator. Comparing Simple Harmonic Motion with Uniform Circular Motion. The Pendulum. Damped Oscillations/ Forced Oscillations. 452 16. Wave Motion Propagation of a Disturbance. Sinusoidal Waves. The Speed of Waves on Strings. Reflection and Transmission. Rate of Energy Transfer by Sinusoidal Waves on Strings. The Linear Wave Equation. 486 17. Sound Waves Speed of Sound Waves. Periodic Sound Waves. Intensity of Periodic Sound Waves. The Doppler Effect. Digital Sound Recording. Motion Picture Sound. 512 18. Superposition and Standing Waves Superposition and Interference. Standing Waves. Standing Waves in a String Fixed at Both Ends. Resonance. Standing Waves in Air Columns. Standing Waves in Rods and Membranes. Beats: Interference in Time. Nonsinusoidal Wave Patterns. 543
    • Part III: THERMODYNAMICS 579 19. Temperature Temperature and the Zeroth Law of Thermodynamics. Thermometers and the Celsius Temperature Scale. The Constant-Volume Gas Thermometer and the Absolute Temperature Scale. Thermal Expansion of Solids and Liquids. Macroscopic Description of an Ideal Gas. 580 20. Heat and the First Law of Thermodynamics Heat and Internal Energy. Specific Heat and Calorimetry. Latent Heat. Work and Heat in Thermodynamic Processes. The First Law of Thermodynamics. Some Applications of the First Law of Thermodynamics. Energy Transfer Mechanisms. 604 21. The Kinetic Theory of Gases Molecular Model of an Ideal Gas. Molar Specific Heat of an Ideal Gas. Adiabatic Processes for an Ideal Gas. The Equipartition of Energy. The Boltzmann Distribution Law. Distribution of Molecular Speeds/ Mean Free Path. 640 22. Heat Engines, Entropy, and the Second Law of Thermodynamics Heat Engines and the Second Law of Thermodynamics. Heat Pumps and Refrigerators. Reversible and Irreversible Processes. The Carnot Engine. Gasoline and Diesel Engines. Entropy. Entropy Changes in Irreversible Processes. Entropy on a Microscopic Scale. 667 Part IV: ELECTRICITY AND MAGNETISM 705 23. Electric Fields Properties of Electric Charges. Charging Objects by Induction. Coulomb's Law. The Electric Field. Electric Field of a Continuous Charge Distribution. Electric Field Lines. Motion of Charged Particles in a Uniform Electric Field. 706 24. Gauss's Law Electric Flux. Gauss's Law. Application of Gauss's Law to Various Charge Distributions. Conductors in Electrostatic Equilibrium. Formal Derivation of Gauss's Law. 739
    • 25. Electric Potential Potential Difference and Electric Potential. Potential Differences in a Uniform Electric Field. Electric Potential and Potential Energy Due to Point Charges. Obtaining the Value of the Electric Field from the Electric Potential. Electric Potential Due to Continuous Charge Distributions. Electric Potential Due to a Charged Conductor. The Millikan Oil-Drop Experiment. Applications of Electrostatics. 762 26. Capacitance and Dielectrics Definition of Capacitance. Calculating Capacitance. Combinations of Capacitors. Energy Stored in a Charged Capacitor. Capacitors with Dielectrics. Electric Dipole in an Electric Field. An Atomic Description of Dielectrics. 795 27. Current and Resistance Electric Current. Resistance. A Model for Electrical Conduction. Resistance and Temperature. Superconductors. Electrical Power. 831 28. Direct Current Circuits Electromotive Force Resistors in Series and Parallel. Kirchhoff's Rules. RC Circuits. Electrical Meters. Household Wiring and Electrical Safety. 858 29. Magnetic Fields Magnetic Field and Forces. Magnetic Force Acting on a Current-Carrying Conductor. Torque on a Current Loop in a Uniform Magnetic Field. Motion of a Charged Particle in a Uniform Magnetic Field. Applications Involving Charged Particles Moving in a Magnetic Field. The Hall Effect. 894 30. Sources of Magnetic Field The Biot-Savart Law. The Magnetic Force Between Two Parallel Conductors. Ampere's Law. The Magnetic Field of a Solenoid. Magnetic Flux. Gauss's Law in Magnetism. Displacement Current and the General Form of Ampere's Law. Magnetism in Matter. The Magnetic Field of the Earth. 926
    • 31. Faraday's Law Faraday's Law of Induction. Motional emf. Lenz's Law. Induced emf and Electric Fields. Generators and Motors/ Eddy Currents. Maxwell's Equations. 967 32. Inductance Self-Inductance. RL Circuits. Energy in a Magnetic Field. Mutual Inductance. Oscillations in an LC Circuit. The RLC Circuit. 1003 33. Alternating Current Circuits AC Sources. Resistors in an AC Circuit. Inductors in an AC Circuit. Capacitors in an AC Circuit. The RLC Series Circuit. Power in an AC Circuit. Resonance in a Series RLC Circuit. The Transformer and Power Transmission. Rectifiers and Filters. 1033 34. Electromagnetic Waves Maxwell's Equations and Hertz's Discoveries. Plane Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure. Production of Electromagnetic Waves by an Antenna. 1066 Part V: LIGHT AND OPTICS 35. The Nature of Light and the Laws of Geometric Optics The Nature of Light. Measurements of the Speed of Light. The Ray Approximation in Geometric Optics. Reflection. Refraction. Huygens's Principle. Dispersion and Prisms. Total Internal Reflection. Fermat's Principle. 1093 1094
    • 36. Image Formation Images Formed by Flat Mirrors. Images Formed by Spherical Mirrors. Images Formed by Refraction. Thin Lenses. Lens Aberrations. The Camera. The Eye. The Simple Magnifier. The Compound Microscope. The Telescope. 1126 37. Interference of Light Waves Conditions for Interference. Young's Double-Slit Experiment. Intensity Distribution of the Double-Slit Interference Pattern. Phasor Addition of Waves. Change of Phase Due to Reflection. Interference in Thin Films. The Michelson Interferometer. 1176 38. Diffraction Patterns and Polarization Introduction to Diffraction Patterns. Diffraction Patterns from Narrow Slits. Resolution of Single-Slit and Circular Apertures. The Diffraction Grating. Diffraction of X-rays by Crystals. Polarization of Light Waves. 1205 Part VI: MODERN PHYSICS 1243 39. Relativity The Principle of Galilean Relativity. The Michelson-Morley Experiment. Einstein's Principle of Relativity. Consequences of the Special Theory of Relativity. The Lorentz Transformation Equations. The Lorentz Velocity Transformation Equations Relativistic Linear Momentum and the Relativistic Form of Newton's Laws. Relativistic Energy. Mass and Energy. The General Theory of Relativity. 1244 APPENDIXES: A. Tables Conversion Factors. Symbols, Dimensions, and Units of Physical Quantities. Table of Atomic Masses. B. Mathematics Review Scientific Notation. Algebra. Geometry. Trigonometry. Series Expansions. Differential Calculus. Integral Calculus. Propagation of Uncertainty. C. Periodic Table of the Elements D. SI Units E. Nobel Prize Winners Answers to Odd-Numbered Problems Index A.1 A.1 A.14 A.30 A.32 A.33 A.37 I.1
    • Mechanics PA R T 1 hysics, the most fundamental physical science, is concerned with the basic principles of the Universe. It is the foundation upon which the other sciences— astronomy, biology, chemistry, and geology—are based. The beauty of physics lies in the simplicity of the fundamental physical theories and in the manner in which just a small number of fundamental concepts, equations, and assumptions can alter and expand our view of the world around us. The study of physics can be divided into six main areas: P 1. classical mechanics, which is concerned with the motion of objects that are large relative to atoms and move at speeds much slower than the speed of light; 2. relativity, which is a theory describing objects moving at any speed, even speeds approaching the speed of light; 3. thermodynamics, which deals with heat, work, temperature, and the statistical behavior of systems with large numbers of particles; 4. electromagnetism, which is concerned with electricity, magnetism, and electromagnetic fields; 5. optics, which is the study of the behavior of light and its interaction with materials; 6. quantum mechanics, a collection of theories connecting the behavior of matter at the submicroscopic level to macroscopic observations. The disciplines of mechanics and electromagnetism are basic to all other branches of classical physics (developed before 1900) and modern physics (c. 1900–present). The first part of this textbook deals with classical mechanics, sometimes referred to as Newtonian mechanics or simply mechanics. This is an appropriate place to begin an introductory text because many of the basic principles used to understand mechanical systems can later be used to describe such natural phenomena as waves and the transfer of energy by heat. Furthermore, the laws of conservation of energy and momentum introduced in mechanics retain their importance in the fundamental theories of other areas of physics. Today, classical mechanics is of vital importance to students from all disciplines. It is highly successful in describing the motions of different objects, such as planets, rockets, and baseballs. In the first part of the text, we shall describe the laws of classical mechanics and examine a wide range of phenomena that can be understood with these fundamental ideas. I ᭣ Liftoff of the space shuttle Columbia. The tragic accident of February 1, 2003 that took the lives of all seven astronauts aboard happened just before Volume 1 of this book went to press. The launch and operation of a space shuttle involves many fundamental principles of classical mechanics, thermodynamics, and electromagnetism. We study the principles of classical mechanics in Part 1 of this text, and apply these principles to rocket propulsion in Chapter 9. (NASA) 1
    • Chapter 1 Physics and Measurement CHAPTE R OUTLI N E 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model Building 1.3 Density and Atomic Mass 1.4 Dimensional Analysis 1.5 Conversion of Units 1.6 Estimates and Order-ofMagnitude Calculations 1.7 Significant Figures L The workings of a mechanical clock. Complicated timepieces have been built for centuries in an effort to measure time accurately. Time is one of the basic quantities that we use in studying the motion of objects. (elektraVision/Index Stock Imagery) 2
    • Like all other sciences, physics is based on experimental observations and quantitative measurements. The main objective of physics is to find the limited number of fundamental laws that govern natural phenomena and to use them to develop theories that can predict the results of future experiments. The fundamental laws used in developing theories are expressed in the language of mathematics, the tool that provides a bridge between theory and experiment. When a discrepancy between theory and experiment arises, new theories must be formulated to remove the discrepancy. Many times a theory is satisfactory only under limited conditions; a more general theory might be satisfactory without such limitations. For example, the laws of motion discovered by Isaac Newton (1642–1727) in the 17th century accurately describe the motion of objects moving at normal speeds but do not apply to objects moving at speeds comparable with the speed of light. In contrast, the special theory of relativity developed by Albert Einstein (1879–1955) in the early 1900s gives the same results as Newton’s laws at low speeds but also correctly describes motion at speeds approaching the speed of light. Hence, Einstein’s special theory of relativity is a more general theory of motion. Classical physics includes the theories, concepts, laws, and experiments in classical mechanics, thermodynamics, optics, and electromagnetism developed before 1900. Important contributions to classical physics were provided by Newton, who developed classical mechanics as a systematic theory and was one of the originators of calculus as a mathematical tool. Major developments in mechanics continued in the 18th century, but the fields of thermodynamics and electricity and magnetism were not developed until the latter part of the 19th century, principally because before that time the apparatus for controlled experiments was either too crude or unavailable. A major revolution in physics, usually referred to as modern physics, began near the end of the 19th century. Modern physics developed mainly because of the discovery that many physical phenomena could not be explained by classical physics. The two most important developments in this modern era were the theories of relativity and quantum mechanics. Einstein’s theory of relativity not only correctly described the motion of objects moving at speeds comparable to the speed of light but also completely revolutionized the traditional concepts of space, time, and energy. The theory of relativity also shows that the speed of light is the upper limit of the speed of an object and that mass and energy are related. Quantum mechanics was formulated by a number of distinguished scientists to provide descriptions of physical phenomena at the atomic level. Scientists continually work at improving our understanding of fundamental laws, and new discoveries are made every day. In many research areas there is a great deal of overlap among physics, chemistry, and biology. Evidence for this overlap is seen in the names of some subspecialties in science—biophysics, biochemistry, chemical physics, biotechnology, and so on. Numerous technological advances in recent times are the result of the efforts of many scientists, engineers, and technicians. Some of the most notable developments in the latter half of the 20th century were (1) unmanned planetary explorations and manned moon landings, (2) microcircuitry and high-speed computers, (3) sophisticated imaging techniques used in scientific research and medicine, and 3
    • 4 C H A P T E R 1 • Physics and Measurement (4) several remarkable results in genetic engineering. The impacts of such developments and discoveries on our society have indeed been great, and it is very likely that future discoveries and developments will be exciting, challenging, and of great benefit to humanity. 1.1 Standards of Length, Mass, and Time The laws of physics are expressed as mathematical relationships among physical quantities that we will introduce and discuss throughout the book. Most of these quantities are derived quantities, in that they can be expressed as combinations of a small number of basic quantities. In mechanics, the three basic quantities are length, mass, and time. All other quantities in mechanics can be expressed in terms of these three. If we are to report the results of a measurement to someone who wishes to reproduce this measurement, a standard must be defined. It would be meaningless if a visitor from another planet were to talk to us about a length of 8 “glitches” if we do not know the meaning of the unit glitch. On the other hand, if someone familiar with our system of measurement reports that a wall is 2 meters high and our unit of length is defined to be 1 meter, we know that the height of the wall is twice our basic length unit. Likewise, if we are told that a person has a mass of 75 kilograms and our unit of mass is defined to be 1 kilogram, then that person is 75 times as massive as our basic unit.1 Whatever is chosen as a standard must be readily accessible and possess some property that can be measured reliably. Measurements taken by different people in different places must yield the same result. In 1960, an international committee established a set of standards for the fundamental quantities of science. It is called the SI (Système International), and its units of length, mass, and time are the meter, kilogram, and second, respectively. Other SI standards established by the committee are those for temperature (the kelvin), electric current (the ampere), luminous intensity (the candela), and the amount of substance (the mole). Length In A.D. 1120 the king of England decreed that the standard of length in his country would be named the yard and would be precisely equal to the distance from the tip of his nose to the end of his outstretched arm. Similarly, the original standard for the foot adopted by the French was the length of the royal foot of King Louis XIV. This standard prevailed until 1799, when the legal standard of length in France became the meter, defined as one ten-millionth the distance from the equator to the North Pole along one particular longitudinal line that passes through Paris. Many other systems for measuring length have been developed over the years, but the advantages of the French system have caused it to prevail in almost all countries and in scientific circles everywhere. As recently as 1960, the length of the meter was defined as the distance between two lines on a specific platinum–iridium bar stored under controlled conditions in France. This standard was abandoned for several reasons, a principal one being that the limited accuracy with which the separation between the lines on the bar can be determined does not meet the current requirements of science and technology. In the 1960s and 1970s, the meter was defined as 1 650 763.73 wavelengths of orange-red light emitted from a krypton-86 lamp. However, in October 1983, the meter (m) was redefined as the distance traveled by light in vacuum during a time of 1/299 792 458 second. In effect, this 1 The need for assigning numerical values to various measured physical quantities was expressed by Lord Kelvin (William Thomson) as follows: “I often say that when you can measure what you are speaking about, and express it in numbers, you should know something about it, but when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind. It may be the beginning of knowledge but you have scarcely in your thoughts advanced to the state of science.”
    • S E C T I O N 1 . 1 • Standards of Length, Mass, and Time Table 1.1 L Approximate Values of Some Measured Lengths Length (m) Distance from the Earth to the most remote known quasar Distance from the Earth to the most remote normal galaxies Distance from the Earth to the nearest large galaxy (M 31, the Andromeda galaxy) Distance from the Sun to the nearest star (Proxima Centauri) One lightyear Mean orbit radius of the Earth about the Sun Mean distance from the Earth to the Moon Distance from the equator to the North Pole Mean radius of the Earth Typical altitude (above the surface) of a satellite orbiting the Earth Length of a football field Length of a housefly Size of smallest dust particles Size of cells of most living organisms Diameter of a hydrogen atom Diameter of an atomic nucleus Diameter of a proton 1.4 ϫ 1026 9 ϫ 1025 2 ϫ 1022 4 ϫ 1016 9.46 ϫ 1015 1.50 ϫ 1011 3.84 ϫ 108 1.00 ϫ 107 6.37 ϫ 106 2 ϫ 105 9.1 ϫ 101 5 ϫ 10Ϫ3 ϳ 10Ϫ4 ϳ 10Ϫ5 ϳ 10Ϫ10 ϳ 10Ϫ14 ϳ 10Ϫ15 5 PITFALL PREVENTION 1.1 No Commas in Numbers with Many Digits We will use the standard scientific notation for numbers with more than three digits, in which groups of three digits are separated by spaces rather than commas. Thus, 10 000 is the same as the common American notation of 10,000. Similarly, ␲ ϭ 3.14159265 is written as 3.141 592 65. Table 1.2 Masses of Various Objects (Approximate Values) Mass (kg) latest definition establishes that the speed of light in vacuum is precisely 299 792 458 meters per second. Table 1.1 lists approximate values of some measured lengths. You should study this table as well as the next two tables and begin to generate an intuition for what is meant by a length of 20 centimeters, for example, or a mass of 100 kilograms or a time interval of 3.2 ϫ 107 seconds. Mass The SI unit of mass, the kilogram (kg), is defined as the mass of a specific platinum–iridium alloy cylinder kept at the International Bureau of Weights and Measures at Sèvres, France. This mass standard was established in 1887 and has not been changed since that time because platinum–iridium is an unusually stable alloy. A duplicate of the Sèvres cylinder is kept at the National Institute of Standards and Technology (NIST) in Gaithersburg, Maryland (Fig. 1.1a). Table 1.2 lists approximate values of the masses of various objects. Observable Universe Milky Way galaxy Sun Earth Moon Shark Human Frog Mosquito Bacterium Hydrogen atom Electron ϳ 1052 ϳ 1042 1.99 ϫ 1030 5.98 ϫ 1024 7.36 ϫ 1022 ϳ 103 ϳ 102 ϳ 10Ϫ1 ϳ 10Ϫ5 ϳ 1 ϫ 10Ϫ15 1.67 ϫ 10Ϫ27 9.11 ϫ 10Ϫ31 Time Before 1960, the standard of time was defined in terms of the mean solar day for the year 1900. (A solar day is the time interval between successive appearances of the Sun at the highest point it reaches in the sky each day.) The second was defined as ΂ ΃΂ ΃΂ ΃ 1 60 1 60 1 24 of a mean solar day. The rotation of the Earth is now known to vary slightly with time, however, and therefore this motion is not a good one to use for defining a time standard. In 1967, the second was redefined to take advantage of the high precision attainable in a device known as an atomic clock (Fig. 1.1b), which uses the characteristic frequency of the cesium-133 atom as the “reference clock.” The second (s) is now defined as 9 192 631 770 times the period of vibration of radiation from the cesium atom.2 2 Period is defined as the time interval needed for one complete vibration. L PITFALL PREVENTION 1.2 Reasonable Values Generating intuition about typical values of quantities is important because when solving problems you must think about your end result and determine if it seems reasonable. If you are calculating the mass of a housefly and arrive at a value of 100 kg, this is unreasonable—there is an error somewhere.
    • C H A P T E R 1 • Physics and Measurement (Courtesy of National Institute of Standards and Technology, U.S. Department of Commerce) 6 (a) (b) Figure 1.1 (a) The National Standard Kilogram No. 20, an accurate copy of the International Standard Kilogram kept at Sèvres, France, is housed under a double bell jar in a vault at the National Institute of Standards and Technology. (b) The nation’s primary time standard is a cesium fountain atomic clock developed at the National Institute of Standards and Technology laboratories in Boulder, Colorado. The clock will neither gain nor lose a second in 20 million years. To keep these atomic clocks—and therefore all common clocks and watches that are set to them—synchronized, it has sometimes been necessary to add leap seconds to our clocks. Since Einstein’s discovery of the linkage between space and time, precise measurement of time intervals requires that we know both the state of motion of the clock used to measure the interval and, in some cases, the location of the clock as well. Otherwise, for example, global positioning system satellites might be unable to pinpoint your location with sufficient accuracy, should you need to be rescued. Approximate values of time intervals are presented in Table 1.3. Table 1.3 Approximate Values of Some Time Intervals Time Interval (s) Age of the Universe Age of the Earth Average age of a college student One year One day (time interval for one revolution of the Earth about its axis) One class period Time interval between normal heartbeats Period of audible sound waves Period of typical radio waves Period of vibration of an atom in a solid Period of visible light waves Duration of a nuclear collision Time interval for light to cross a proton 5 ϫ 1017 1.3 ϫ 1017 6.3 ϫ 108 3.2 ϫ 107 8.6 ϫ 104 3.0 ϫ 103 8 ϫ 10Ϫ1 ϳ 10Ϫ3 ϳ 10Ϫ6 ϳ 10Ϫ13 ϳ 10Ϫ15 ϳ 10Ϫ22 ϳ 10Ϫ24
    • S E C T I O N 1 . 2 • Matter and Model Building Table 1.4 Prefixes for Powers of Ten Power Prefix Abbreviation 10Ϫ24 10Ϫ21 10Ϫ18 10Ϫ15 10Ϫ12 10Ϫ9 10Ϫ6 10Ϫ3 10Ϫ2 10Ϫ1 103 106 109 1012 1015 1018 1021 1024 yocto zepto atto femto pico nano micro milli centi deci kilo mega giga tera peta exa zetta yotta y z a f p n ␮ m c d k M G T P E Z Y In addition to SI, another system of units, the U.S. customary system, is still used in the United States despite acceptance of SI by the rest of the world. In this system, the units of length, mass, and time are the foot (ft), slug, and second, respectively. In this text we shall use SI units because they are almost universally accepted in science and industry. We shall make some limited use of U.S. customary units in the study of classical mechanics. In addition to the basic SI units of meter, kilogram, and second, we can also use other units, such as millimeters and nanoseconds, where the prefixes milli- and nanodenote multipliers of the basic units based on various powers of ten. Prefixes for the various powers of ten and their abbreviations are listed in Table 1.4. For example, 10Ϫ 3 m is equivalent to 1 millimeter (mm), and 103 m corresponds to 1 kilometer (km). Likewise, 1 kilogram (kg) is 103 grams (g), and 1 megavolt (MV) is 106 volts (V). 1.2 Matter and Model Building If physicists cannot interact with some phenomenon directly, they often imagine a model for a physical system that is related to the phenomenon. In this context, a model is a system of physical components, such as electrons and protons in an atom. Once we have identified the physical components, we make predictions about the behavior of the system, based on the interactions among the components of the system and/or the interaction between the system and the environment outside the system. As an example, consider the behavior of matter. A 1-kg cube of solid gold, such as that at the left of Figure 1.2, has a length of 3.73 cm on a side. Is this cube nothing but wall-to-wall gold, with no empty space? If the cube is cut in half, the two pieces still retain their chemical identity as solid gold. But what if the pieces are cut again and again, indefinitely? Will the smaller and smaller pieces always be gold? Questions such as these can be traced back to early Greek philosophers. Two of them—Leucippus and his student Democritus—could not accept the idea that such cuttings could go on forever. They speculated that the process ultimately must end when it produces a particle 7
    • 8 C H A P T E R 1 • Physics and Measurement Quark composition of a proton u u d Neutron Gold nucleus Nucleus Proton Gold cube Gold atoms Figure 1.2 Levels of organization in matter. Ordinary matter consists of atoms, and at the center of each atom is a compact nucleus consisting of protons and neutrons. Protons and neutrons are composed of quarks. The quark composition of a proton is shown. that can no longer be cut. In Greek, atomos means “not sliceable.” From this comes our English word atom. Let us review briefly a number of historical models of the structure of matter. The Greek model of the structure of matter was that all ordinary matter consists of atoms, as suggested to the lower right of the cube in Figure 1.2. Beyond that, no additional structure was specified in the model— atoms acted as small particles that interacted with each other, but internal structure of the atom was not a part of the model. In 1897, J. J. Thomson identified the electron as a charged particle and as a constituent of the atom. This led to the first model of the atom that contained internal structure. We shall discuss this model in Chapter 42. Following the discovery of the nucleus in 1911, a model was developed in which each atom is made up of electrons surrounding a central nucleus. A nucleus is shown in Figure 1.2. This model leads, however, to a new question—does the nucleus have structure? That is, is the nucleus a single particle or a collection of particles? The exact composition of the nucleus is not known completely even today, but by the early 1930s a model evolved that helped us understand how the nucleus behaves. Specifically, scientists determined that occupying the nucleus are two basic entities, protons and neutrons. The proton carries a positive electric charge, and a specific chemical element is identified by the number of protons in its nucleus. This number is called the atomic number of the element. For instance, the nucleus of a hydrogen atom contains one proton (and so the atomic number of hydrogen is 1), the nucleus of a helium atom contains two protons (atomic number 2), and the nucleus of a uranium atom contains 92 protons (atomic number 92). In addition to atomic number, there is a second number characterizing atoms—mass number, defined as the number of protons plus neutrons in a nucleus. The atomic number of an element never varies (i.e., the number of protons does not vary) but the mass number can vary (i.e., the number of neutrons varies). The existence of neutrons was verified conclusively in 1932. A neutron has no charge and a mass that is about equal to that of a proton. One of its primary purposes
    • S E C T I O N 1 . 3 • Density and Atomic Mass is to act as a “glue” that holds the nucleus together. If neutrons were not present in the nucleus, the repulsive force between the positively charged particles would cause the nucleus to come apart. But is this where the process of breaking down stops? Protons, neutrons, and a host of other exotic particles are now known to be composed of six different varieties of particles called quarks, which have been given the names of up, down, strange, charmed, 2 bottom, and top. The up, charmed, and top quarks have electric charges of ϩ 3 that of the proton, whereas the down, strange, and bottom quarks have charges of Ϫ 1 that 3 of the proton. The proton consists of two up quarks and one down quark, as shown at the top in Figure 1.2. You can easily show that this structure predicts the correct charge for the proton. Likewise, the neutron consists of two down quarks and one up quark, giving a net charge of zero. This process of building models is one that you should develop as you study physics. You will be challenged with many mathematical problems to solve in this study. One of the most important techniques is to build a model for the problem—identify a system of physical components for the problem, and make predictions of the behavior of the system based on the interactions among the components of the system and/or the interaction between the system and its surrounding environment. 1.3 Density and Atomic Mass In Section 1.1, we explored three basic quantities in mechanics. Let us look now at an example of a derived quantity—density. The density ␳ (Greek letter rho) of any substance is defined as its mass per unit volume: ␳ϵ m V (1.1) For example, aluminum has a density of 2.70 g/cm3, and lead has a density of 11.3 g/cm3. Therefore, a piece of aluminum of volume 10.0 cm3 has a mass of 27.0 g, whereas an equivalent volume of lead has a mass of 113 g. A list of densities for various substances is given in Table 1.5. The numbers of protons and neutrons in the nucleus of an atom of an element are related to the atomic mass of the element, which is defined as the mass of a single atom of the element measured in atomic mass units (u) where 1 u ϭ 1.660 538 7 ϫ 10Ϫ27 kg. Table 1.5 Densities of Various Substances Substance Density ␳ (103 kg/m3) Platinum Gold Uranium Lead Copper Iron Aluminum Magnesium Water Air at atmospheric pressure 21.45 19.3 18.7 11.3 8.92 7.86 2.70 1.75 1.00 0.0012 A table of the letters in the Greek alphabet is provided on the back endsheet of the textbook. 9
    • C H A P T E R 1 • Physics and Measurement 10 The atomic mass of lead is 207 u and that of aluminum is 27.0 u. However, the ratio of atomic masses, 207 u/27.0 u ϭ 7.67, does not correspond to the ratio of densities, (11.3 ϫ 103 kg/m3)/(2.70 ϫ 103 kg/m3) ϭ 4.19. This discrepancy is due to the difference in atomic spacings and atomic arrangements in the crystal structures of the two elements. Quick Quiz 1.1 In a machine shop, two cams are produced, one of aluminum and one of iron. Both cams have the same mass. Which cam is larger? (a) the aluminum cam (b) the iron cam (c) Both cams have the same size. Example 1.1 How Many Atoms in the Cube? A solid cube of aluminum (density 2.70 g/cm3) has a volume of 0.200 cm3. It is known that 27.0 g of aluminum contains 6.02 ϫ 1023 atoms. How many aluminum atoms are contained in the cube? Solution Because density equals mass per unit volume, the mass of the cube is m ϭ ␳V ϭ (2.70 g/cm3)(0.200 cm3) ϭ 0.540 g To solve this problem, we will set up a ratio based on the fact that the mass of a sample of material is proportional to the number of atoms contained in the sample. This technique of solving by ratios is very powerful and should be studied and understood so that it can be applied in future problem solving. Let us express our proportionality as m ϭ kN, where m is the mass of the sample, N is the number of atoms in the sample, and k is an unknown proportionality constant. We L PITFALL PREVENTION 1.3 Setting Up Ratios When using ratios to solve a problem, keep in mind that ratios come from equations. If you start from equations known to be correct and can divide one equation by the other as in Example 1.1 to obtain a useful ratio, you will avoid reasoning errors. So write the known equations first! write this relationship twice, once for the actual sample of aluminum in the problem and once for a 27.0-g sample, and then we divide the first equation by the second: m sample ϭ kN sample m 27.0 g ϭ kN27.0 g : m sample N ϭ sample m 27.0 g N27.0 g Notice that the unknown proportionality constant k cancels, so we do not need to know its value. We now substitute the values: 0.540 g N sample ϭ 27.0 g 6.02 ϫ 1023 atoms Nsample ϭ (0.540 g)(6.02 ϫ 1023 atoms) 27.0 g ϭ 1.20 ϫ 1022 atoms 1.4 Dimensional Analysis The word dimension has a special meaning in physics. It denotes the physical nature of a quantity. Whether a distance is measured in units of feet or meters or fathoms, it is still a distance. We say its dimension is length. The symbols we use in this book to specify the dimensions of length, mass, and time are L, M, and T, respectively.3 We shall often use brackets [ ] to denote the dimensions of a physical quantity. For example, the symbol we use for speed in this book is v, and in our notation the dimensions of speed are written [v] ϭ L/T. As another example, the dimensions of area A are [A] ϭ L2. The dimensions and units of area, volume, speed, and acceleration are listed in Table 1.6. The dimensions of other quantities, such as force and energy, will be described as they are introduced in the text. In many situations, you may have to derive or check a specific equation. A useful and powerful procedure called dimensional analysis can be used to assist in the derivation or to check your final expression. Dimensional analysis makes use of the fact that 3 The dimensions of a quantity will be symbolized by a capitalized, non-italic letter, such as L. The symbol for the quantity itself will be italicized, such as L for the length of an object, or t for time.
    • S E C T I O N 1 . 4 • Dimensional Analysis Table 1.6 Units of Area, Volume, Velocity, Speed, and Acceleration System Area (L2) Volume (L3) Speed (L/T) Acceleration (L/T 2) SI U.S. customary m2 ft2 m3 ft3 m/s ft/s m/s2 ft/s2 dimensions can be treated as algebraic quantities. For example, quantities can be added or subtracted only if they have the same dimensions. Furthermore, the terms on both sides of an equation must have the same dimensions. By following these simple rules, you can use dimensional analysis to help determine whether an expression has the correct form. The relationship can be correct only if the dimensions on both sides of the equation are the same. To illustrate this procedure, suppose you wish to derive an equation for the position x of a car at a time t if the car starts from rest and moves with constant acceleration a. In Chapter 2, we shall find that the correct expression is x ϭ 1 at 2. Let us use 2 dimensional analysis to check the validity of this expression. The quantity x on the left side has the dimension of length. For the equation to be dimensionally correct, the quantity on the right side must also have the dimension of length. We can perform a dimensional check by substituting the dimensions for acceleration, L/T2 (Table 1.6), and time, T, into the equation. That is, the dimensional form of the equation x ϭ 1 at 2 is 2 Lϭ L и T2 ϭ L T2 The dimensions of time cancel as shown, leaving the dimension of length on the righthand side. A more general procedure using dimensional analysis is to set up an expression of the form x ϰ ant m where n and m are exponents that must be determined and the symbol ϰ indicates a proportionality. This relationship is correct only if the dimensions of both sides are the same. Because the dimension of the left side is length, the dimension of the right side must also be length. That is, [antm] ϭ L ϭ L1T0 Because the dimensions of acceleration are L/T2 and the dimension of time is T, we have (L/T2)n Tm ϭ L1T0 (Ln Tm Ϫ2n) ϭ L1T0 The exponents of L and T must be the same on both sides of the equation. From the exponents of L, we see immediately that n ϭ 1. From the exponents of T, we see that m Ϫ 2n ϭ 0, which, once we substitute for n, gives us m ϭ 2. Returning to our original expression x ϰ antm, we conclude that x ϰ at 2. This result differs by a factor of 1 from 2 the correct expression, which is x ϭ 1 at 2. 2 Quick Quiz 1.2 True or False: Dimensional analysis can give you the numerical value of constants of proportionality that may appear in an algebraic expression. L 11 PITFALL PREVENTION 1.4 Symbols for Quantities Some quantities have a small number of symbols that represent them. For example, the symbol for time is almost always t. Others quantities might have various symbols depending on the usage. Length may be described with symbols such as x, y, and z (for position), r (for radius), a, b, and c (for the legs of a right triangle), ᐉ (for the length of an object), d (for a distance), h (for a height), etc.
    • C H A P T E R 1 • Physics and Measurement 12 Example 1.2 Analysis of an Equation Show that the expression v ϭ at is dimensionally correct, where v represents speed, a acceleration, and t an instant of time. The same table gives us L/T2 for the dimensions of acceleration, and so the dimensions of at are [at] ϭ Solution For the speed term, we have from Table 1.6 Therefore, the expression is dimensionally correct. (If the expression were given as v ϭ at 2 it would be dimensionally incorrect. Try it and see!) L [v] ϭ T Example 1.3 Analysis of a Power Law Suppose we are told that the acceleration a of a particle moving with uniform speed v in a circle of radius r is proportional to some power of r, say r n, and some power of v, say vm. Determine the values of n and m and write the simplest form of an equation for the acceleration. Solution Let us take a to be This dimensional equation is balanced under the conditions n ϩ mϭ 1 where k is a dimensionless constant of proportionality. Knowing the dimensions of a, r, and v, we see that the dimensional equation must be ΂ ΃ L L ϭ Ln T2 T PITFALL PREVENTION 1.5 Always Include Units When performing calculations, include the units for every quantity and carry the units through the entire calculation. Avoid the temptation to drop the units early and then attach the expected units once you have an answer. By including the units in every step, you can detect errors if the units for the answer turn out to be incorrect. m ϭ Lnϩm Tm and mϭ 2 Therefore n ϭ Ϫ 1, and we can write the acceleration expression as a ϭ kr nvm L L L 2 Tϭ T T a ϭ kr Ϫ1v 2 ϭ k v2 r When we discuss uniform circular motion later, we shall see that k ϭ 1 if a consistent set of units is used. The constant k would not equal 1 if, for example, v were in km/h and you wanted a in m/s2. 1.5 Conversion of Units Sometimes it is necessary to convert units from one measurement system to another, or to convert within a system, for example, from kilometers to meters. Equalities between SI and U.S. customary units of length are as follows: 1 mile ϭ 1 609 m ϭ 1.609 km 1 m ϭ 39.37 in. ϭ 3.281 ft 1 ft ϭ 0.304 8 m ϭ 30.48 cm 1 in. ϭ 0.025 4 m ϭ 2.54 cm (exactly) A more complete list of conversion factors can be found in Appendix A. Units can be treated as algebraic quantities that can cancel each other. For example, suppose we wish to convert 15.0 in. to centimeters. Because 1 in. is defined as exactly 2.54 cm, we find that 15.0 in. ϭ (15.0 in.)΂ 2.54 cm ΃ ϭ 38.1 cm 1 in. where the ratio in parentheses is equal to 1. Notice that we choose to put the unit of an inch in the denominator and it cancels with the unit in the original quantity. The remaining unit is the centimeter, which is our desired result. Quick Quiz 1.3 The distance between two cities is 100 mi. The number of kilometers between the two cities is (a) smaller than 100 (b) larger than 100 (c) equal to 100.
    • S E C T I O N 1 . 6 • Estimates and Order-of-Magnitude Calculations Example 1.4 Is He Speeding? On an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 38.0 m/s. Is this car exceeding the speed limit of 75.0 mi/h? Figure 1.3 shows the speedometer of an automobile, with speeds in both mi/h and km/h. Can you check the conversion we just performed using this photograph? Solution We first convert meters to miles: (38.0 m/s) ΂ 1 1 mim ΃ ϭ 2.36 ϫ 10 609 Ϫ2 mi/s Now we convert seconds to hours: (2.36 ϫ 10 Ϫ2 mi/s) ΂ 160 s ΃ ΂ 601min ΃ ϭ 85.0 mi/h min h What If? What if the driver is from outside the U.S. and is familiar with speeds measured in km/h? What is the speed of the car in km/h? Answer We can convert our final answer to the appropriate units: (85.0 mi/h) ΂ 1.609 km 1 mi ΃ Phil Boorman/Getty Images Thus, the car is exceeding the speed limit and should slow down. Figure 1.3 The speedometer of a vehicle that shows speeds in both miles per hour and kilometers per hour. ϭ 137 km/h 1.6 Estimates and Order-of-Magnitude Calculations It is often useful to compute an approximate answer to a given physical problem even when little information is available. This answer can then be used to determine whether or not a more precise calculation is necessary. Such an approximation is usually based on certain assumptions, which must be modified if greater precision is needed. We will sometimes refer to an order of magnitude of a certain quantity as the power of ten of the number that describes that quantity. Usually, when an order-ofmagnitude calculation is made, the results are reliable to within about a factor of 10. If a quantity increases in value by three orders of magnitude, this means that its value increases by a factor of about 103 ϭ 1 000. We use the symbol ϳ for “is on the order of.” Thus, 0.008 6 ϳ 10Ϫ2 0.002 1 ϳ 10Ϫ3 720 ϳ 103 The spirit of order-of-magnitude calculations, sometimes referred to as “guesstimates” or “ball-park figures,” is given in the following quotation: “Make an estimate before every calculation, try a simple physical argument . . . before every derivation, guess the answer to every puzzle.”4 Inaccuracies caused by guessing too low for one number are often canceled out by other guesses that are too high. You will find that with practice your guesstimates become better and better. Estimation problems can be fun to work as you freely drop digits, venture reasonable approximations for 4 13 E. Taylor and J. A. Wheeler, Spacetime Physics: Introduction to Special Relativity, 2nd ed., San Francisco, W. H. Freeman & Company, Publishers, 1992, p. 20.
    • 14 C H A P T E R 1 • Physics and Measurement unknown numbers, make simplifying assumptions, and turn the question around into something you can answer in your head or with minimal mathematical manipulation on paper. Because of the simplicity of these types of calculations, they can be performed on a small piece of paper, so these estimates are often called “back-of-theenvelope calculations.” Example 1.5 Breaths in a Lifetime Estimate the number of breaths taken during an average life span. Solution We start by guessing that the typical life span is about 70 years. The only other estimate we must make in this example is the average number of breaths that a person takes in 1 min. This number varies, depending on whether the person is exercising, sleeping, angry, serene, and so forth. To the nearest order of magnitude, we shall choose 10 breaths per minute as our estimate of the average. (This is certainly closer to the true value than 1 breath per minute or 100 breaths per minute.) The number of minutes in a year is approximately 1 yr days ΂ 400 yr ΃ ΂ 125 h ΃ ΂ 601min ΃ ϭ 6 ϫ 10 1 day h 5 min Notice how much simpler it is in the expression above to multiply 400 ϫ 25 than it is to work with the more accurate 365 ϫ 24. These approximate values for the number of days Example 1.6 in a year and the number of hours in a day are close enough for our purposes. Thus, in 70 years there will be (70 yr)(6 ϫ 105 min/yr) ϭ 4 ϫ 107 min. At a rate of 10 breaths/min, an individual would take 4 ϫ 108 breaths in a lifetime, or on the order of 109 breaths. What If? What if the average life span were estimated as 80 years instead of 70? Would this change our final estimate? Answer We could claim that (80 yr)(6 ϫ 105 min/yr) ϭ 5 ϫ 107 min, so that our final estimate should be 5 ϫ 108 breaths. This is still on the order of 109 breaths, so an orderof-magnitude estimate would be unchanged. Furthermore, 80 years is 14% larger than 70 years, but we have overestimated the total time interval by using 400 days in a year instead of 365 and 25 hours in a day instead of 24. These two numbers together result in an overestimate of 14%, which cancels the effect of the increased life span! It’s a Long Way to San Jose Estimate the number of steps a person would take walking from New York to Los Angeles. Now we switch to scientific notation so that we can do the calculation mentally: Solution Without looking up the distance between these two cities, you might remember from a geography class that they are about 3 000 mi apart. The next approximation we must make is the length of one step. Of course, this length depends on the person doing the walking, but we can estimate that each step covers about 2 ft. With our estimated step size, we can determine the number of steps in 1 mi. Because this is a rough calculation, we round 5 280 ft/mi to 5 000 ft/mi. (What percentage error does this introduce?) This conversion factor gives us (3 ϫ 103 mi)(2.5 ϫ 103 steps/mi) ϭ 7.5 ϫ 106 steps ϳ 107 steps So if we intend to walk across the United States, it will take us on the order of ten million steps. This estimate is almost certainly too small because we have not accounted for curving roads and going up and down hills and mountains. Nonetheless, it is probably within an order of magnitude of the correct answer. 5 000 ft/mi ϭ 2 500 steps/mi 2 ft/step Example 1.7 How Much Gas Do We Use? Estimate the number of gallons of gasoline used each year by all the cars in the United States. Solution Because there are about 280 million people in the United States, an estimate of the number of cars in the country is 100 million (guessing that there are between two and three people per car). We also estimate that the average distance each car travels per year is 10 000 mi. If we assume a gasoline consumption of 20 mi/gal or 0.05 gal/mi, then each car uses about 500 gal/yr. Multiplying this by the total number of cars in the United States gives an estimated total consumption of 5 ϫ 1010 gal ϳ 1011 gal.
    • S E C T I O N 1 . 7 • Significant Figures 1.7 15 Significant Figures When certain quantities are measured, the measured values are known only to within the limits of the experimental uncertainty. The value of this uncertainty can depend on various factors, such as the quality of the apparatus, the skill of the experimenter, and the number of measurements performed. The number of significant figures in a measurement can be used to express something about the uncertainty. As an example of significant figures, suppose that we are asked in a laboratory experiment to measure the area of a computer disk label using a meter stick as a measuring instrument. Let us assume that the accuracy to which we can measure the length of the label is Ϯ 0.1 cm. If the length is measured to be 5.5 cm, we can claim only that its length lies somewhere between 5.4 cm and 5.6 cm. In this case, we say that the measured value has two significant figures. Note that the significant figures include the first estimated digit. Likewise, if the label’s width is measured to be 6.4 cm, the actual value lies between 6.3 cm and 6.5 cm. Thus we could write the measured values as (5.5 Ϯ 0.1) cm and (6.4 Ϯ 0.1) cm. Now suppose we want to find the area of the label by multiplying the two measured values. If we were to claim the area is (5.5 cm)(6.4 cm) ϭ 35.2 cm2, our answer would be unjustifiable because it contains three significant figures, which is greater than the number of significant figures in either of the measured quantities. A good rule of thumb to use in determining the number of significant figures that can be claimed in a multiplication or a division is as follows: When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the quantity having the lowest number of significant figures. The same rule applies to division. Applying this rule to the previous multiplication example, we see that the answer for the area can have only two significant figures because our measured quantities have only two significant figures. Thus, all we can claim is that the area is 35 cm2, realizing that the value can range between (5.4 cm)(6.3 cm) ϭ 34 cm2 and (5.6 cm)(6.5 cm) ϭ 36 cm2. Zeros may or may not be significant figures. Those used to position the decimal point in such numbers as 0.03 and 0.007 5 are not significant. Thus, there are one and two significant figures, respectively, in these two values. When the zeros come after other digits, however, there is the possibility of misinterpretation. For example, suppose the mass of an object is given as 1 500 g. This value is ambiguous because we do not know whether the last two zeros are being used to locate the decimal point or whether they represent significant figures in the measurement. To remove this ambiguity, it is common to use scientific notation to indicate the number of significant figures. In this case, we would express the mass as 1.5 ϫ 103 g if there are two significant figures in the measured value, 1.50 ϫ 103 g if there are three significant figures, and 1.500 ϫ 103 g if there are four. The same rule holds for numbers less than 1, so that 2.3 ϫ 10Ϫ4 has two significant figures (and so could be written 0.000 23) and 2.30 ϫ 10Ϫ4 has three significant figures (also written 0.000 230). In general, a significant figure in a measurement is a reliably known digit (other than a zero used to locate the decimal point) or the first estimated digit. For addition and subtraction, you must consider the number of decimal places when you are determining how many significant figures to report: When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum. L PITFALL PREVENTION 1.6 Read Carefully Notice that the rule for addition and subtraction is different from that for multiplication and division. For addition and subtraction, the important consideration is the number of decimal places, not the number of significant figures.
    • 16 C H A P T E R 1 • Physics and Measurement For example, if we wish to compute 123 ϩ 5.35, the answer is 128 and not 128.35. If we compute the sum 1.000 1 ϩ 0.000 3 ϭ 1.000 4, the result has five significant figures, even though one of the terms in the sum, 0.000 3, has only one significant figure. Likewise, if we perform the subtraction 1.002 Ϫ 0.998 ϭ 0.004, the result has only one significant figure even though one term has four significant figures and the other has three. In this book, most of the numerical examples and end-of-chapter problems will yield answers having three significant figures. When carrying out estimates we shall typically work with a single significant figure. If the number of significant figures in the result of an addition or subtraction must be reduced, there is a general rule for rounding off numbers, which states that the last digit retained is to be increased by 1 if the last digit dropped is greater than 5. If the last digit dropped is less than 5, the last digit retained remains as it is. If the last digit dropped is equal to 5, the remaining digit should be rounded to the nearest even number. (This helps avoid accumulation of errors in long arithmetic processes.) A technique for avoiding error accumulation is to delay rounding of numbers in a long calculation until you have the final result. Wait until you are ready to copy the final answer from your calculator before rounding to the correct number of significant figures. Quick Quiz 1.4 Suppose you measure the position of a chair with a meter stick and record that the center of the seat is 1.043 860 564 2 m from a wall. What would a reader conclude from this recorded measurement? Example 1.8 Installing a Carpet A carpet is to be installed in a room whose length is measured to be 12.71 m and whose width is measured to be 3.46 m. Find the area of the room. Solution If you multiply 12.71 m by 3.46 m on your calculator, you will see an answer of 43.976 6 m2. How many of these numbers should you claim? Our rule of thumb for multiplication tells us that you can claim only the number of significant figures in your answer as are present in the measured quantity having the lowest number of significant figures. In this example, the lowest number of significant figures is three in 3.46 m, so we should express our final answer as 44.0 m2. S U M MARY Take a practice test for this chapter by clicking on the Practice Test link at http://www.pse6.com. The three fundamental physical quantities of mechanics are length, mass, and time, which in the SI system have the units meters (m), kilograms (kg), and seconds (s), respectively. Prefixes indicating various powers of ten are used with these three basic units. The density of a substance is defined as its mass per unit volume. Different substances have different densities mainly because of differences in their atomic masses and atomic arrangements. The method of dimensional analysis is very powerful in solving physics problems. Dimensions can be treated as algebraic quantities. By making estimates and performing order-of-magnitude calculations, you should be able to approximate the answer to a problem when there is not enough information available to completely specify an exact solution. When you compute a result from several measured numbers, each of which has a certain accuracy, you should give the result with the correct number of significant figures. When multiplying several quantities, the number of significant figures in the
    • Problems 17 final answer is the same as the number of significant figures in the quantity having the lowest number of significant figures. The same rule applies to division. When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum. QU ESTIONS 1. What types of natural phenomena could serve as time standards? 2. Suppose that the three fundamental standards of the metric system were length, density, and time rather than length, mass, and time. The standard of density in this system is to be defined as that of water. What considerations about water would you need to address to make sure that the standard of density is as accurate as possible? 3. The height of a horse is sometimes given in units of “hands.” Why is this a poor standard of length? 4. Express the following quantities using the prefixes given in Table 1.4: (a) 3 ϫ 10Ϫ4 m (b) 5 ϫ 10Ϫ5 s (c) 72 ϫ 102 g. 5. Suppose that two quantities A and B have different dimensions. Determine which of the following arithmetic operations could be physically meaningful: (a) A ϩ B (b) A/B (c) B Ϫ A (d) AB. 6. If an equation is dimensionally correct, does this mean that the equation must be true? If an equation is not dimensionally correct, does this mean that the equation cannot be true? 7. Do an order-of-magnitude calculation for an everyday situation you encounter. For example, how far do you walk or drive each day? 8. Find the order of magnitude of your age in seconds. 9. What level of precision is implied in an order-of-magnitude calculation? 10. Estimate the mass of this textbook in kilograms. If a scale is available, check your estimate. 11. In reply to a student’s question, a guard in a natural history museum says of the fossils near his station, “When I started work here twenty-four years ago, they were eighty million years old, so you can add it up.” What should the student conclude about the age of the fossils? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide = coached solution with hints available at http://www.pse6.com = computer useful in solving problem = paired numerical and symbolic problems Section 1.2 Matter and Model Building Note: Consult the endpapers, appendices, and tables in the text whenever necessary in solving problems. For this chapter, Appendix B.3 may be particularly useful. Answers to odd-numbered problems appear in the back of the book. L d (a) 1. A crystalline solid consists of atoms stacked up in a repeating lattice structure. Consider a crystal as shown in Figure P1.1a. The atoms reside at the corners of cubes of side L ϭ 0.200 nm. One piece of evidence for the regular arrangement of atoms comes from the flat surfaces along which a crystal separates, or cleaves, when it is broken. Suppose this crystal cleaves along a face diagonal, as shown in Figure P1.1b. Calculate the spacing d between two adjacent atomic planes that separate when the crystal cleaves. (b) Figure P1.1
    • 18 C H A P T E R 1 • Physics and Measurement Section 1.3 Density and Atomic Mass 2. Use information on the endpapers of this book to calculate the average density of the Earth. Where does the value fit among those listed in Tables 1.5 and 14.1? Look up the density of a typical surface rock like granite in another source and compare the density of the Earth to it. mass of a section 1.50 m long? (b) Assume that the atoms are predominantly iron, with atomic mass 55.9 u. How many atoms are in this section? 15.0 cm 3. The standard kilogram is a platinum–iridium cylinder 39.0 mm in height and 39.0 mm in diameter. What is the density of the material? 4. A major motor company displays a die-cast model of its first automobile, made from 9.35 kg of iron. To celebrate its hundredth year in business, a worker will recast the model in gold from the original dies. What mass of gold is needed to make the new model? 5. What mass of a material with density ␳ is required to make a hollow spherical shell having inner radius r 1 and outer radius r 2? 6. Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is five times greater. Find its radius. 7. Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in grams. The atomic masses of these atoms are 4.00 u, 55.9 u, and 207 u, respectively. 8. The paragraph preceding Example 1.1 in the text mentions that the atomic mass of aluminum is 27.0 u ϭ 27.0 ϫ 1.66 ϫ 10Ϫ27 kg. Example 1.1 says that 27.0 g of aluminum contains 6.02 ϫ 1023 atoms. (a) Prove that each one of these two statements implies the other. (b) What If ? What if it’s not aluminum? Let M represent the numerical value of the mass of one atom of any chemical element in atomic mass units. Prove that M grams of the substance contains a particular number of atoms, the same number for all elements. Calculate this number precisely from the value for u quoted in the text. The number of atoms in M grams of an element is called Avogadro’s number NA. The idea can be extended: Avogadro’s number of molecules of a chemical compound has a mass of M grams, where M atomic mass units is the mass of one molecule. Avogadro’s number of atoms or molecules is called one mole, symbolized as 1 mol. A periodic table of the elements, as in Appendix C, and the chemical formula for a compound contain enough information to find the molar mass of the compound. (c) Calculate the mass of one mole of water, H2O. (d) Find the molar mass of CO2. 1.00 cm 36.0 cm 1.00 cm Figure P1.11 12. A child at the beach digs a hole in the sand and uses a pail to fill it with water having a mass of 1.20 kg. The mass of one molecule of water is 18.0 u. (a) Find the number of water molecules in this pail of water. (b) Suppose the quantity of water on Earth is constant at 1.32 ϫ 1021 kg. How many of the water molecules in this pail of water are likely to have been in an equal quantity of water that once filled one particular claw print left by a Tyrannosaur hunting on a similar beach? Section 1.4 Dimensional Analysis 13. The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position s ϭ ka mt n, where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if m ϭ 1 and n ϭ 2. Can this analysis give the value of k? 14. Figure P1.14 shows a frustrum of a cone. Of the following mensuration (geometrical) expressions, which describes (a) the total circumference of the flat circular faces (b) the volume (c) the area of the curved surface? (i) ␲(r 1 ϩ r 2)[h2 ϩ (r 1 Ϫ r 2)2]1/2 (ii) 2␲(r 1 ϩ r 2) (iii) ␲h(r 12 ϩ r 1r 2 ϩ r 22). 9. On your wedding day your lover gives you a gold ring of mass 3.80 g. Fifty years later its mass is 3.35 g. On the average, how many atoms were abraded from the ring during each second of your marriage? The atomic mass of gold is 197 u. 10. A small cube of iron is observed under a microscope. The edge of the cube is 5.00 ϫ 10Ϫ6 cm long. Find (a) the mass of the cube and (b) the number of iron atoms in the cube. The atomic mass of iron is 55.9 u, and its density is 7.86 g/cm3. 11. A structural I beam is made of steel. A view of its crosssection and its dimensions are shown in Figure P1.11. The density of the steel is 7.56 ϫ 103 kg/m3. (a) What is the r1 h r2 Figure P1.14
    • Problems 16. (a) A fundamental law of motion states that the acceleration of an object is directly proportional to the resultant force exerted on the object and inversely proportional to its mass. If the proportionality constant is defined to have no dimensions, determine the dimensions of force. (b) The newton is the SI unit of force. According to the results for (a), how can you express a force having units of newtons using the fundamental units of mass, length, and time? 17. Newton’s law of universal gravitation is represented by Fϭ GMm r2 Here F is the magnitude of the gravitational force exerted by one small object on another, M and m are the masses of the objects, and r is a distance. Force has the SI units kg · m/s2. What are the SI units of the proportionality constant G ? Section 1.5 Conversion of Units 18. A worker is to paint the walls of a square room 8.00 ft high and 12.0 ft along each side. What surface area in square meters must she cover? 19. Suppose your hair grows at the rate 1/32 in. per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly layers of atoms are assembled in this protein synthesis. 20. The volume of a wallet is 8.50 in.3 Convert this value to m3, using the definition 1 in. ϭ 2.54 cm. 21. A rectangular building lot is 100 ft by 150 ft. Determine the area of this lot in m2. 22. An auditorium measures 40.0 m ϫ 20.0 m ϫ 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic feet and (b) the weight of air in the room in pounds? 23. Assume that it takes 7.00 minutes to fill a 30.0-gal gasoline tank. (a) Calculate the rate at which the tank is filled in gallons per second. (b) Calculate the rate at which the tank is filled in cubic meters per second. (c) Determine the time interval, in hours, required to fill a 1-m3 volume at the same rate. (1 U.S. gal ϭ 231 in.3) 24. Find the height or length of these natural wonders in kilometers, meters and centimeters. (a) The longest cave system in the world is the Mammoth Cave system in central Kentucky. It has a mapped length of 348 mi. (b) In the United States, the waterfall with the greatest single drop is Ribbon Falls, which falls 1 612 ft. (c) Mount McKinley in Denali National Park, Alaska, is America’s highest mountain at a height of 20 320 ft. (d) The deepest canyon in the United States is King’s Canyon in California with a depth of 8 200 ft. 25. A solid piece of lead has a mass of 23.94 g and a volume of 2.10 cm3. From these data, calculate the density of lead in SI units (kg/m3). 26. A section of land has an area of 1 square mile and contains 640 acres. Determine the number of square meters in 1 acre. 27. An ore loader moves 1 200 tons/h from a mine to the surface. Convert this rate to lb/s, using 1 ton ϭ 2 000 lb. 28. (a) Find a conversion factor to convert from miles per hour to kilometers per hour. (b) In the past, a federal law mandated that highway speed limits would be 55 mi/h. Use the conversion factor of part (a) to find this speed in kilometers per hour. (c) The maximum highway speed is now 65 mi/h in some places. In kilometers per hour, how much increase is this over the 55 mi/h limit? 29. At the time of this book’s printing, the U.S. national debt is about $6 trillion. (a) If payments were made at the rate of $1 000 per second, how many years would it take to pay off the debt, assuming no interest were charged? (b) A dollar bill is about 15.5 cm long. If six trillion dollar bills were laid end to end around the Earth’s equator, how many times would they encircle the planet? Take the radius of the Earth at the equator to be 6 378 km. (Note: Before doing any of these calculations, try to guess at the answers. You may be very surprised.) 30. The mass of the Sun is 1.99 ϫ 1030 kg, and the mass of an atom of hydrogen, of which the Sun is mostly composed, is 1.67 ϫ 10Ϫ27 kg. How many atoms are in the Sun? 31. One gallon of paint (volume ϭ 3.78 ϫ 10Ϫ3 m3) covers an area of 25.0 m2. What is the thickness of the paint on the wall? 32. A pyramid has a height of 481 ft and its base covers an area of 13.0 acres (Fig. P1.32). If the volume of a pyramid is given by the expression V ϭ 1 Bh, where B is the area of 3 the base and h is the height, find the volume of this pyramid in cubic meters. (1 acre ϭ 43 560 ft2) Sylvain Grandadam/Photo Researchers, Inc. 15. Which of the following equations are dimensionally correct? (a) vf ϭ vi ϩ ax (b) y ϭ (2 m)cos(kx), where k ϭ 2 mϪ1. 19 Figure P1.32 Problems 32 and 33. 33. The pyramid described in Problem 32 contains approximately 2 million stone blocks that average 2.50 tons each. Find the weight of this pyramid in pounds. 34. Assuming that 70% of the Earth’s surface is covered with water at an average depth of 2.3 mi, estimate the mass of the water on the Earth in kilograms. 35. A hydrogen atom has a diameter of approximately 1.06 ϫ 10Ϫ10 m, as defined by the diameter of the spherical electron cloud around the nucleus. The hydrogen nucleus has a diameter of approximately 2.40 ϫ 10Ϫ15 m. (a) For a scale model, represent the diameter of the hydrogen atom by the length of an American football field
    • 20 C H A P T E R 1 • Physics and Measurement (100 yd ϭ 300 ft), and determine the diameter of the nucleus in millimeters. (b) The atom is how many times larger in volume than its nucleus? How many tons of aluminum does this represent? In your solution state the quantities you measure or estimate and the values you take for them. 36. The nearest stars to the Sun are in the Alpha Centauri multiple-star system, about 4.0 ϫ 1013 km away. If the Sun, with a diameter of 1.4 ϫ 109 m, and Alpha Centauri A are both represented by cherry pits 7.0 mm in diameter, how far apart should the pits be placed to represent the Sun and its neighbor to scale? 47. To an order of magnitude, how many piano tuners are in New York City? The physicist Enrico Fermi was famous for asking questions like this on oral Ph.D. qualifying examinations. His own facility in making order-of-magnitude calculations is exemplified in Problem 45.48. 37. The diameter of our disk-shaped galaxy, the Milky Way, is about 1.0 ϫ 105 lightyears (ly). The distance to Messier 31, which is Andromeda, the spiral galaxy nearest to the Milky Way, is about 2.0 million ly. If a scale model represents the Milky Way and Andromeda galaxies as dinner plates 25 cm in diameter, determine the distance between the two plates. 38. The mean radius of the Earth is 6.37 ϫ 106 m, and that of the Moon is 1.74 ϫ 108 cm. From these data calculate (a) the ratio of the Earth’s surface area to that of the Moon and (b) the ratio of the Earth’s volume to that of the Moon. Recall that the surface area of a sphere is 4␲r 2 and the volume of a sphere is 4 ␲r 3. 3 39. One cubic meter (1.00 m3) of aluminum has a mass of 2.70 ϫ 103 kg, and 1.00 m3 of iron has a mass of 7.86 ϫ 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 2.00 cm on an equal-arm balance. 40. Let ␳Al represent the density of aluminum and ␳ Fe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius r Fe on an equal-arm balance. Section 1.6 Estimates and Order-of-Magnitude Calculations 41. Estimate the number of Ping-Pong balls that would fit into a typical-size room (without being crushed). In your solution state the quantities you measure or estimate and the values you take for them. 42. An automobile tire is rated to last for 50 000 miles. To an order of magnitude, through how many revolutions will it turn? In your solution state the quantities you measure or estimate and the values you take for them. 43. Grass grows densely everywhere on a quarter-acre plot of land. What is the order of magnitude of the number of blades of grass on this plot? Explain your reasoning. Note that 1 acre ϭ 43 560 ft2. Section 1.7 Significant Figures Note: Appendix B.8 on propagation of uncertainty may be useful in solving some problems in this section. 48. A rectangular plate has a length of (21.3 Ϯ 0.2) cm and a width of (9.8 Ϯ 0.1) cm. Calculate the area of the plate, including its uncertainty. 49. The radius of a circle is measured to be (10.5 Ϯ 0.2) m. Calculate the (a) area and (b) circumference of the circle and give the uncertainty in each value. 50. How many significant figures are in the following numbers? (a) 78.9 Ϯ 0.2 (b) 3.788 ϫ 109 (c) 2.46 ϫ 10Ϫ6 (d) 0.005 3. 51. The radius of a solid sphere is measured to be (6.50 Ϯ 0.20) cm, and its mass is measured to be (1.85 Ϯ 0.02) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density. 52. Carry out the following arithmetic operations: (a) the sum of the measured values 756, 37.2, 0.83, and 2.5; (b) the product 0.003 2 ϫ 356.3; (c) the product 5.620 ϫ ␲. 53. The tropical year, the time from vernal equinox to the next vernal equinox, is the basis for our calendar. It contains 365.242 199 days. Find the number of seconds in a tropical year. 54. A farmer measures the distance around a rectangular field. The length of the long sides of the rectangle is found to be 38.44 m, and the length of the short sides is found to be 19.5 m. What is the total distance around the field? 55. A sidewalk is to be constructed around a swimming pool that measures (10.0 Ϯ 0.1) m by (17.0 Ϯ 0.1) m. If the sidewalk is to measure (1.00 Ϯ 0.01) m wide by (9.0 Ϯ 0.1) cm thick, what volume of concrete is needed, and what is the approximate uncertainty of this volume? 44. Approximately how many raindrops fall on a one-acre lot during a one-inch rainfall? Explain your reasoning. 45. Compute the order of magnitude of the mass of a bathtub half full of water. Compute the order of magnitude of the mass of a bathtub half full of pennies. In your solution list the quantities you take as data and the value you measure or estimate for each. 46. Soft drinks are commonly sold in aluminum containers. To an order of magnitude, how many such containers are thrown away or recycled each year by U.S. consumers? Additional Problems 56. In a situation where data are known to three significant digits, we write 6.379 m ϭ 6.38 m and 6.374 m ϭ 6.37 m. When a number ends in 5, we arbitrarily choose to write 6.375 m ϭ 6.38 m. We could equally well write 6.375 m ϭ 6.37 m, “rounding down” instead of “rounding up,” because we would change the number 6.375 by equal increments in both cases. Now consider an order-of-magnitude
    • Problems estimate, in which we consider factors rather than increments. We write 500 m ϳ 103 m because 500 differs from 100 by a factor of 5 while it differs from 1 000 by only a factor of 2. We write 437 m ϳ 103 m and 305 m ϳ 102 m. What distance differs from 100 m and from 1 000 m by equal factors, so that we could equally well choose to represent its order of magnitude either as ϳ 102 m or as ϳ 103 m? 57. For many electronic applications, such as in computer chips, it is desirable to make components as small as possible to keep the temperature of the components low and to increase the speed of the device. Thin metallic coatings (films) can be used instead of wires to make electrical connections. Gold is especially useful because it does not oxidize readily. Its atomic mass is 197 u. A gold film can be no thinner than the size of a gold atom. Calculate the minimum coating thickness, assuming that a gold atom occupies a cubical volume in the film that is equal to the volume it occupies in a large piece of metal. This geometric model yields a result of the correct order of magnitude. 58. The basic function of the carburetor of an automobile is to “atomize” the gasoline and mix it with air to promote rapid combustion. As an example, assume that 30.0 cm3 of gasoline is atomized into N spherical droplets, each with a radius of 2.00 ϫ 10Ϫ5 m. What is the total surface area of these N spherical droplets? 59. The consumption of natural gas by a company satisfies the empirical equation V ϭ 1.50t ϩ 0.008 00t 2, where V is the volume in millions of cubic feet and t the time in months. Express this equation in units of cubic feet and seconds. Assign proper units to the coefficients. Assume a month is equal to 30.0 days. 60. In physics it is important to use mathematical approximations. Demonstrate that for small angles (Ͻ 20°) tan ␣ Ϸ sin ␣ Ϸ ␣ ϭ ␲␣Ј/180° where ␣ is in radians and ␣Ј is in degrees. Use a calculator to find the largest angle for which tan ␣ may be approximated by sin ␣ if the error is to be less than 10.0%. 61. A high fountain of water is located at the center of a circular pool as in Figure P1.61. Not wishing to get his feet wet, 21 a student walks around the pool and measures its circumference to be 15.0 m. Next, the student stands at the edge of the pool and uses a protractor to gauge the angle of elevation of the top of the fountain to be 55.0°. How high is the fountain? 62. Collectible coins are sometimes plated with gold to enhance their beauty and value. Consider a commemorative quarter-dollar advertised for sale at $4.98. It has a diameter of 24.1 mm, a thickness of 1.78 mm, and is completely covered with a layer of pure gold 0.180 ␮m thick. The volume of the plating is equal to the thickness of the layer times the area to which it is applied. The patterns on the faces of the coin and the grooves on its edge have a negligible effect on its area. Assume that the price of gold is $10.0 per gram. Find the cost of the gold added to the coin. Does the cost of the gold significantly enhance the value of the coin? 63. There are nearly ␲ ϫ 107 s in one year. Find the percentage error in this approximation, where “percentage error’’ is defined as Percentage error ϭ & assumed value Ϫ true value & ϫ 100% true value 64. Assume that an object covers an area A and has a uniform height h. If its cross-sectional area is uniform over its height, then its volume is given by V ϭ Ah. (a) Show that V ϭ Ah is dimensionally correct. (b) Show that the volumes of a cylinder and of a rectangular box can be written in the form V ϭ Ah, identifying A in each case. (Note that A, sometimes called the “footprint” of the object, can have any shape and the height can be replaced by average thickness in general.) 65. A child loves to watch as you fill a transparent plastic bottle with shampoo. Every horizontal cross-section is a circle, but the diameters of the circles have different values, so that the bottle is much wider in some places than others. You pour in bright green shampoo with constant volume flow rate 16.5 cm3/s. At what rate is its level in the bottle rising (a) at a point where the diameter of the bottle is 6.30 cm and (b) at a point where the diameter is 1.35 cm? 66. One cubic centimeter of water has a mass of 1.00 ϫ 10Ϫ3 kg. (a) Determine the mass of 1.00 m3 of water. (b) Biological substances are 98% water. Assume that they have the same density as water to estimate the masses of a cell that has a diameter of 1.0 ␮m, a human kidney, and a fly. Model the kidney as a sphere with a radius of 4.0 cm and the fly as a cylinder 4.0 mm long and 2.0 mm in diameter. 55.0˚ Figure P1.61 67. Assume there are 100 million passenger cars in the United States and that the average fuel consumption is 20 mi/gal of gasoline. If the average distance traveled by each car is 10 000 mi/yr, how much gasoline would be saved per year if average fuel consumption could be increased to 25 mi/gal? 68. A creature moves at a speed of 5.00 furlongs per fortnight (not a very common unit of speed). Given that 1 furlong ϭ 220 yards and 1 fortnight ϭ 14 days, determine the speed of the creature in m/s. What kind of creature do you think it might be?
    • 22 C H A P T E R 1 • Physics and Measurement 69. The distance from the Sun to the nearest star is about 4 ϫ 1016 m. The Milky Way galaxy is roughly a disk of diameter ϳ 1021 m and thickness ϳ 1019 m. Find the order of magnitude of the number of stars in the Milky Way. Assume the distance between the Sun and our nearest neighbor is typical. 70. The data in the following table represent measurements of the masses and dimensions of solid cylinders of aluminum, copper, brass, tin, and iron. Use these data to calculate the densities of these substances. Compare your results for aluminum, copper, and iron with those given in Table 1.5. Substance Mass (g) Diameter (cm) Length (cm) Aluminum Copper Brass Tin Iron 51.5 56.3 94.4 69.1 216.1 2.52 1.23 1.54 1.75 1.89 3.75 5.06 5.69 3.74 9.77 71. (a) How many seconds are in a year? (b) If one micrometeorite (a sphere with a diameter of 1.00 ϫ 10Ϫ 6 m) strikes each square meter of the Moon each second, how many years will it take to cover the Moon to a depth of 1.00 m? To solve this problem, you can consider a cubic box on the Moon 1.00 m on each edge, and find how long it will take to fill the box. Answers to Quick Quizzes 1.1 (a). Because the density of aluminum is smaller than that of iron, a larger volume of aluminum is required for a given mass than iron. 1.2 False. Dimensional analysis gives the units of the proportionality constant but provides no information about its numerical value. To determine its numerical value requires either experimental data or geometrical reasoning. For example, in the generation of the equation x ϭ 1 at 2, because the factor 1 is dimensionless, there is 2 2 no way of determining it using dimensional analysis. 1.3 (b). Because kilometers are shorter than miles, a larger number of kilometers is required for a given distance than miles. 1.4 Reporting all these digits implies you have determined the location of the center of the chair’s seat to the nearest Ϯ 0.000 000 000 1 m. This roughly corresponds to being able to count the atoms in your meter stick because each of them is about that size! It would be better to record the measurement as 1.044 m: this indicates that you know the position to the nearest millimeter, assuming the meter stick has millimeter markings on its scale.
    • Chapter 2 Motion in One Dimension CHAPTE R OUTLI N E 2.1 Position, Velocity, and Speed 2.2 Instantaneous Velocity and Speed 2.3 Acceleration 2.4 Motion Diagrams 2.5 One-Dimensional Motion with Constant Acceleration 2.6 Freely Falling Objects 2.7 Kinematic Equations Derived from Calculus General Problem-Solving Strategy L One of the physical quantities we will study in this chapter is the velocity of an object moving in a straight line. Downhill skiers can reach velocities with a magnitude greater than 100 km/h. (Jean Y. Ruszniewski/Getty Images) 23
    • As a first step in studying classical mechanics, we describe motion in terms of space and time while ignoring the agents that caused that motion. This portion of classical mechanics is called kinematics. (The word kinematics has the same root as cinema. Can you see why?) In this chapter we consider only motion in one dimension, that is, motion along a straight line. We first define position, displacement, velocity, and acceleration. Then, using these concepts, we study the motion of objects traveling in one dimension with a constant acceleration. From everyday experience we recognize that motion represents a continuous change in the position of an object. In physics we can categorize motion into three types: translational, rotational, and vibrational. A car moving down a highway is an example of translational motion, the Earth’s spin on its axis is an example of rotational motion, and the back-and-forth movement of a pendulum is an example of vibrational motion. In this and the next few chapters, we are concerned only with translational motion. (Later in the book we shall discuss rotational and vibrational motions.) In our study of translational motion, we use what is called the particle model— we describe the moving object as a particle regardless of its size. In general, a particle is a point-like object—that is, an object with mass but having infinitesimal size. For example, if we wish to describe the motion of the Earth around the Sun, we can treat the Earth as a particle and obtain reasonably accurate data about its orbit. This approximation is justified because the radius of the Earth’s orbit is large compared with the dimensions of the Earth and the Sun. As an example on a much smaller scale, it is possible to explain the pressure exerted by a gas on the walls of a container by treating the gas molecules as particles, without regard for the internal structure of the molecules. 2.1 Position, Velocity, and Speed Position 24 The motion of a particle is completely known if the particle’s position in space is known at all times. A particle’s position is the location of the particle with respect to a chosen reference point that we can consider to be the origin of a coordinate system. Consider a car moving back and forth along the x axis as in Figure 2.1a. When we begin collecting position data, the car is 30 m to the right of a road sign, which we will use to identify the reference position x ϭ 0. (Let us assume that all data in this example are known to two significant figures. To convey this information, we should report the initial position as 3.0 ϫ 10 1 m. We have written this value in the simpler form 30 m to make the discussion easier to follow.) We will use the particle model by identifying some point on the car, perhaps the front door handle, as a particle representing the entire car. We start our clock and once every 10 s note the car’s position relative to the sign at x ϭ 0. As you can see from Table 2.1, the car moves to the right (which we have
    • S E C T I O N 2 . 1 • Position, Velocity, and Speed 25 Table 2.1 –60 –50 Position of the Car at Various Times IT LIM /h 30 km –40 –30 –20 Ꭽ –10 0 10 Position 20 30 40 ൵ 50 60 x(m) ൴ –60 –50 ൳ –40 –30 –20 ∆x Ꭿ –10 0 10 20 30 –20 ൴ –40 ൵ t(s) 0 50 60 x(m) Active Figure 2.1 (a) A car moves back and forth along a straight line taken to be the x axis. Because we are interested only in the car’s translational motion, we can model it as a particle. (b) Position–time graph for the motion of the “particle.” ൳ 0 –60 40 Ꭿ ∆t 20 30 52 38 0 Ϫ37 Ϫ53 Ꭾ 40 Ꭽ 0 10 20 30 40 50 IT LIM /h 30 km (a) x(m) 60 x(m) Ꭽ Ꭾ Ꭿ ൳ ൴ ൵ Ꭾ t(s) 10 20 30 40 50 (b) At the Active Figures link at http://www.pse6.com, you can move each of the six points Ꭽ through ൵ and observe the motion of the car pictorially and graphically as it follows a smooth path through the six points. defined as the positive direction) during the first 10 s of motion, from position Ꭽ to position Ꭾ. After Ꭾ, the position values begin to decrease, suggesting that the car is backing up from position Ꭾ through position ൵. In fact, at ൳, 30 s after we start measuring, the car is alongside the road sign (see Figure 2.1a) that we are using to mark our origin of coordinates. It continues moving to the left and is more than 50 m to the left of the sign when we stop recording information after our sixth data point. A graphical representation of this information is presented in Figure 2.1b. Such a plot is called a position–time graph. Given the data in Table 2.1, we can easily determine the change in position of the car for various time intervals. The displacement of a particle is defined as its change in position in some time interval. As it moves from an initial position xi to a final position xf , the displacement of the particle is given by xf Ϫ x i . We use the Greek letter delta (⌬) to denote the change in a quantity. Therefore, we write the displacement, or change in position, of the particle as ⌬x ϵ x f Ϫ x i (2.1) Displacement
    • C H A P T E R 2 • Motion in One Dimension From this definition we see that ⌬x is positive if xf is greater than xi and negative if xf is less than xi. It is very important to recognize the difference between displacement and distance traveled. Distance is the length of a path followed by a particle. Consider, for example, the basketball players in Figure 2.2. If a player runs from his own basket down the court to the other team’s basket and then returns to his own basket, the displacement of the player during this time interval is zero, because he ended up at the same point as he started. During this time interval, however, he covered a distance of twice the length of the basketball court. Displacement is an example of a vector quantity. Many other physical quantities, including position, velocity, and acceleration, also are vectors. In general, a vector quantity requires the specification of both direction and magnitude. By contrast, a scalar quantity has a numerical value and no direction. In this chapter, we use positive (ϩ) and negative (Ϫ) signs to indicate vector direction. We can do this because the chapter deals with one-dimensional motion only; this means that any object we study can be moving only along a straight line. For example, for horizontal motion let us arbitrarily specify to the right as being the positive direction. It follows that any object always moving to the right undergoes a positive displacement ⌬x Ͼ 0, and any object moving to the left undergoes a negative displacement, so that ⌬x Ͻ 0. We shall treat vector quantities in greater detail in Chapter 3. For our basketball player in Figure 2.2, if the trip from his own basket to the opposing basket is described by a displacement of ϩ 28 m, the trip in the reverse direction represents a displacement of Ϫ 28 m. Each trip, however, represents a distance of 28 m, because distance is a scalar quantity. The total distance for the trip down the court and back is 56 m. Distance, therefore, is always represented as a positive number, while displacement can be either positive or negative. There is one very important point that has not yet been mentioned. Note that the data in Table 2.1 results only in the six data points in the graph in Figure 2.1b. The smooth curve drawn through the six points in the graph is only a possibility of the actual motion of the car. We only have information about six instants of time—we have no idea what happened in between the data points. The smooth curve is a guess as to what happened, but keep in mind that it is only a guess. If the smooth curve does represent the actual motion of the car, the graph contains information about the entire 50-s interval during which we watch the car move. It is much easier to see changes in position from the graph than from a verbal description or even a table of numbers. For example, it is clear that the car was covering more ground during the middle of the 50-s interval than at the end. Between positions Ꭿ and ൳, the car traveled almost 40 m, but during the last 10 s, between positions ൴ and ൵, it moved less than half that far. A common way of comparing these different motions is to divide the displacement ⌬x that occurs between two clock readings by the length of that particular time interval ⌬t. This turns out to be a very useful ratio, one that we shall use many times. This ratio has been given a special – name—average velocity. The average velocity v x of a particle is defined as the Ken White/Allsport/Getty Images 26 Figure 2.2 On this basketball court, players run back and forth for the entire game. The distance that the players run over the duration of the game is nonzero. The displacement of the players over the duration of the game is approximately zero because they keep returning to the same point over and over again.
    • S E C T I O N 2 . 1 • Position, Velocity, and Speed 27 particle’s displacement ∆x divided by the time interval ∆t during which that displacement occurs: vx ϵ ⌬x ⌬t (2.2) Average velocity where the subscript x indicates motion along the x axis. From this definition we see that average velocity has dimensions of length divided by time (L/T)—meters per second in SI units. The average velocity of a particle moving in one dimension can be positive or negative, depending on the sign of the displacement. (The time interval ⌬t is always positive.) If the coordinate of the particle increases in time (that is, if xf Ͼ x i ), then ⌬x is positive and vx ϭ ⌬x/⌬t is positive. This case corresponds to a particle moving in the positive x direction, that is, toward larger values of x. If the coordinate decreases in time (that is, if xf Ͻ x i ) then ⌬x is negative and hence vx is negative. This case corresponds to a particle moving in the negative x direction. We can interpret average velocity geometrically by drawing a straight line between any two points on the position–time graph in Figure 2.1b. This line forms the hypotenuse of a right triangle of height ⌬x and base ⌬t. The slope of this line is the ratio ⌬x/⌬t, which is what we have defined as average velocity in Equation 2.2. For example, the line between positions Ꭽ and Ꭾ in Figure 2.1b has a slope equal to the average velocity of the car between those two times, (52 m Ϫ 30 m)/(10 s Ϫ 0) ϭ 2.2 m/s. In everyday usage, the terms speed and velocity are interchangeable. In physics, however, there is a clear distinction between these two quantities. Consider a marathon runner who runs more than 40 km, yet ends up at his starting point. His total displacement is zero, so his average velocity is zero! Nonetheless, we need to be able to quantify how fast he was running. A slightly different ratio accomplishes this for us. The average speed of a particle, a scalar quantity, is defined as the total distance traveled divided by the total time interval required to travel that distance: Average speed ϭ total distance total time (2.3) The SI unit of average speed is the same as the unit of average velocity: meters per second. However, unlike average velocity, average speed has no direction and hence carries no algebraic sign. Notice the distinction between average velocity and average speed—average velocity (Eq. 2.2) is the displacement divided by the time interval, while average speed (Eq. 2.3) is the distance divided by the time interval. Knowledge of the average velocity or average speed of a particle does not provide information about the details of the trip. For example, suppose it takes you 45.0 s to travel 100 m down a long straight hallway toward your departure gate at an airport. At the 100-m mark, you realize you missed the rest room, and you return back 25.0 m along the same hallway, taking 10.0 s to make the return trip. The magnitude of the average velocity for your trip is ϩ 75.0 m/55.0 s ϭ ϩ 1.36 m/s. The average speed for your trip is 125 m/55.0 s ϭ 2.27 m/s. You may have traveled at various speeds during the walk. Neither average velocity nor average speed provides information about these details. Quick Quiz 2.1 Under which of the following conditions is the magnitude of the average velocity of a particle moving in one dimension smaller than the average speed over some time interval? (a) A particle moves in the ϩ x direction without reversing. (b) A particle moves in the Ϫ x direction without reversing. (c) A particle moves in the ϩ x direction and then reverses the direction of its motion. (d) There are no conditions for which this is true. Average speed L PITFALL PREVENTION 2.1 Average Speed and Average Velocity The magnitude of the average velocity is not the average speed. For example, consider the marathon runner discussed here. The magnitude of the average velocity is zero, but the average speed is clearly not zero.
    • C H A P T E R 2 • Motion in One Dimension 28 Example 2.1 Calculating the Average Velocity and Speed Find the displacement, average velocity, and average speed of the car in Figure 2.1a between positions Ꭽ and ൵. Solution From the position–time graph given in Figure 2.1b, note that xA ϭ 30 m at tA ϭ 0 s and that x F ϭ Ϫ 53 m at t F ϭ 50 s. Using these values along with the definition of displacement, Equation 2.1, we find that ⌬x ϭ x F Ϫ x A ϭ Ϫ53 m Ϫ 30 m ϭ Ϫ83 m This result means that the car ends up 83 m in the negative direction (to the left, in this case) from where it started. This number has the correct units and is of the same order of magnitude as the supplied data. A quick look at Figure 2.1a indicates that this is the correct answer. It is difficult to estimate the average velocity without completing the calculation, but we expect the units to be meters per second. Because the car ends up to the left of where we started taking data, we know the average velocity must be negative. From Equation 2.2, vx ϭ ⌬x ⌬t ϭ xf Ϫ xi tf Ϫ ti ϭ xF Ϫ xA tF Ϫ tA Ϫ53 m Ϫ 30 m Ϫ83 m ϭ ϭ 50 s Ϫ 0 s 50 s ϭ Ϫ1.7 m/s We cannot unambiguously find the average speed of the car from the data in Table 2.1, because we do not have information about the positions of the car between the data points. If we adopt the assumption that the details of the car’s position are described by the curve in Figure 2.1b, then the distance traveled is 22 m (from Ꭽ to Ꭾ) plus 105 m (from Ꭾ to ൵) for a total of 127 m. We find the car’s average speed for this trip by dividing the distance by the total time (Eq. 2.3): Average speed ϭ 127 m ϭ 2.5 m/s 50 s 2.2 Instantaneous Velocity and Speed L PITFALL PREVENTION 2.2 Slopes of Graphs In any graph of physical data, the slope represents the ratio of the change in the quantity represented on the vertical axis to the change in the quantity represented on the horizontal axis. Remember that a slope has units (unless both axes have the same units). The units of slope in Figure 2.1b and Figure 2.3 are m/s, the units of velocity. Often we need to know the velocity of a particle at a particular instant in time, rather than the average velocity over a finite time interval. For example, even though you might want to calculate your average velocity during a long automobile trip, you would be especially interested in knowing your velocity at the instant you noticed the police car parked alongside the road ahead of you. In other words, you would like to be able to specify your velocity just as precisely as you can specify your position by noting what is happening at a specific clock reading—that is, at some specific instant. It may not be immediately obvious how to do this. What does it mean to talk about how fast something is moving if we “freeze time” and talk only about an individual instant? This is a subtle point not thoroughly understood until the late 1600s. At that time, with the invention of calculus, scientists began to understand how to describe an object’s motion at any moment in time. To see how this is done, consider Figure 2.3a, which is a reproduction of the graph in Figure 2.1b. We have already discussed the average velocity for the interval during which the car moved from position Ꭽ to position Ꭾ (given by the slope of the dark blue line) and for the interval during which it moved from Ꭽ to ൵ (represented by the slope of the light blue line and calculated in Example 2.1). Which of these two lines do you think is a closer approximation of the initial velocity of the car? The car starts out by moving to the right, which we defined to be the positive direction. Therefore, being positive, the value of the average velocity during the Ꭽ to Ꭾ interval is more representative of the initial value than is the value of the average velocity during the Ꭽ to ൵ interval, which we determined to be negative in Example 2.1. Now let us focus on the dark blue line and slide point Ꭾ to the left along the curve, toward point Ꭽ, as in Figure 2.3b. The line between the points becomes steeper and steeper, and as the two points become extremely close together, the line becomes a tangent line to the curve, indicated by the green line in Figure 2.3b. The slope of this tangent line
    • S E C T I O N 2 . 2 • Instantaneous Velocity and Speed 60 x(m) 60 Ꭾ Ꭿ 40 Ꭽ Ꭾ 20 ᎮᎮ Ꭾ ൳ 0 40 –20 ൴ –40 –60 29 ൵ 0 10 20 30 (a) 40 50 t(s) Ꭽ (b) Active Figure 2.3 (a) Graph representing the motion of the car in Figure 2.1. (b) An enlargement of the upper-left-hand corner of the graph shows how the blue line between positions Ꭽ and Ꭾ approaches the green tangent line as point Ꭾ is moved closer to point Ꭽ. At the Active Figures link at http://www.pse6.com, you can move point Ꭾ as suggested in (b) and observe the blue line approaching the green tangent line. represents the velocity of the car at the moment we started taking data, at point Ꭽ. What we have done is determine the instantaneous velocity at that moment. In other words, the instantaneous velocity vx equals the limiting value of the ratio ⌬xր⌬t as ⌬t approaches zero:1 vx ϵ lim ⌬t : 0 ⌬x ⌬t (2.4) In calculus notation, this limit is called the derivative of x with respect to t, written dx/dt: vx ϵ lim ⌬t : 0 ⌬x dx ϭ ⌬t dt (2.5) The instantaneous velocity can be positive, negative, or zero. When the slope of the position–time graph is positive, such as at any time during the first 10 s in Figure 2.3, vx is positive—the car is moving toward larger values of x. After point Ꭾ, vx is negative because the slope is negative—the car is moving toward smaller values of x. At point Ꭾ, the slope and the instantaneous velocity are zero—the car is momentarily at rest. From here on, we use the word velocity to designate instantaneous velocity. When it is average velocity we are interested in, we shall always use the adjective average. The instantaneous speed of a particle is defined as the magnitude of its instantaneous velocity. As with average speed, instantaneous speed has no direction associated with it and hence carries no algebraic sign. For example, if one particle has an instantaneous velocity of ϩ 25 m/s along a given line and another particle has an instantaneous velocity of Ϫ 25 m/s along the same line, both have a speed2 of 25 m/s. 1 Note that the displacement ⌬x also approaches zero as ⌬t approaches zero, so that the ratio looks like 0/0. As ⌬x and ⌬t become smaller and smaller, the ratio ⌬x/⌬t approaches a value equal to the slope of the line tangent to the x-versus-t curve. 2 As with velocity, we drop the adjective for instantaneous speed: “Speed” means instantaneous speed. Instantaneous velocity L PITFALL PREVENTION 2.3 Instantaneous Speed and Instantaneous Velocity In Pitfall Prevention 2.1, we argued that the magnitude of the average velocity is not the average speed. Notice the difference when discussing instantaneous values. The magnitude of the instantaneous velocity is the instantaneous speed. In an infinitesimal time interval, the magnitude of the displacement is equal to the distance traveled by the particle.
    • 30 C H A P T E R 2 • Motion in One Dimension Conceptual Example 2.2 The Velocity of Different Objects Consider the following one-dimensional motions: (A) A ball thrown directly upward rises to a highest point and falls back into the thrower’s hand. (B) A race car starts from rest and speeds up to 100 m/s. (C) A spacecraft drifts through space at constant velocity. Are there any points in the motion of these objects at which the instantaneous velocity has the same value as the average velocity over the entire motion? If so, identify the point(s). Solution (A) The average velocity for the thrown ball is zero because the ball returns to the starting point; thus its displacement is zero. (Remember that average velocity is defined as ⌬x/⌬t.) There is one point at which the instantaneous velocity is zero—at the top of the motion. Example 2.3 (B) The car’s average velocity cannot be evaluated unambiguously with the information given, but it must be some value between 0 and 100 m/s. Because the car will have every instantaneous velocity between 0 and 100 m/s at some time during the interval, there must be some instant at which the instantaneous velocity is equal to the average velocity. (C) Because the spacecraft’s instantaneous velocity is constant, its instantaneous velocity at any time and its average velocity over any time interval are the same. Average and Instantaneous Velocity A particle moves along the x axis. Its position varies with time according to the expression x ϭ Ϫ 4t ϩ 2t 2 where x is in meters and t is in seconds.3 The position–time graph for this motion is shown in Figure 2.4. Note that the particle moves in the negative x direction for the first second of motion, is momentarily at rest at the moment t ϭ 1 s, and moves in the positive x direction at times t Ͼ 1 s. x(m) 10 8 6 4 (A) Determine the displacement of the particle in the time intervals t ϭ 0 to t ϭ 1 s and t ϭ 1 s to t ϭ 3 s. Solution During the first time interval, the slope is negative and hence the average velocity is negative. Thus, we know that the displacement between Ꭽ and Ꭾ must be a negative number having units of meters. Similarly, we expect the displacement between Ꭾ and ൳ to be positive. In the first time interval, we set ti ϭ tA ϭ 0 and tf ϭ t B ϭ 1 s. Using Equation 2.1, with x ϭ Ϫ 4t ϩ 2t 2, we obtain for the displacement between t ϭ 0 and t ϭ 1 s, ⌬xA : B ϭ xf Ϫ xi ϭ x B Ϫ xA 0 To calculate the displacement during the second time interval (t ϭ 1 s to t ϭ 3 s), we set ti ϭ t B ϭ 1 s and tf ϭ t D ϭ 3 s: ⌬x B : D ϭ x f Ϫ x i ϭ x D Ϫ x B ϭ [Ϫ4(3) ϩ 2(3)2] Ϫ [Ϫ4(1) ϩ 2(1)2] ϭ ϩ8 m Solution In the first time interval, ⌬t ϭ tf Ϫ ti ϭ t B Ϫ t A ϭ 1 s. Therefore, using Equation 2.2 and the displacement calculated in (a), we find that vx(A : B) ϭ ⌬x A : B Ϫ2 m ϭ ϭ Ϫ2 m/s ⌬t 1s t(s) Ꭾ 0 1 2 3 4 Figure 2.4 (Example 2.3) Position–time graph for a particle having an x coordinate that varies in time according to the expression x ϭ Ϫ 4t ϩ 2t 2. In the second time interval, ⌬t ϭ 2 s; therefore, vx(B : D) ϭ ⌬x B : D 8m ϭ ϭ ϩ4 m/s ⌬t 2s These values are the same as the slopes of the lines joining these points in Figure 2.4. (C) Find the instantaneous velocity of the particle at t ϭ 2.5 s. Solution We can guess that this instantaneous velocity must be of the same order of magnitude as our previous results, that is, a few meters per second. By measuring the slope of the green line at t ϭ 2.5 s in Figure 2.4, we find that These displacements can also be read directly from the position–time graph. (B) Calculate the average velocity during these two time intervals. Ꭿ Ꭽ –2 –4 ൳ Slope = –2 m/s 2 ϭ [Ϫ4(1) ϩ 2(1)2] Ϫ [Ϫ4(0) ϩ 2(0)2] ϭ Ϫ2 m Slope = 4 m/s vx ϭ ϩ6 m/s 3 Simply to make it easier to read, we write the expression as x ϭ Ϫ 4t ϩ 2t 2 rather than as x ϭ (Ϫ 4.00 m/s)t ϩ (2.00 m/s2)t 2.00. When an equation summarizes measurements, consider its coefficients to have as many significant digits as other data quoted in a problem. Consider its coefficients to have the units required for dimensional consistency. When we start our clocks at t ϭ 0, we usually do not mean to limit the precision to a single digit. Consider any zero value in this book to have as many significant figures as you need.
    • S E C T I O N 2 . 3 • Acceleration 2.3 31 Acceleration In the last example, we worked with a situation in which the velocity of a particle changes while the particle is moving. This is an extremely common occurrence. (How constant is your velocity as you ride a city bus or drive on city streets?) It is possible to quantify changes in velocity as a function of time similarly to the way in which we quantify changes in position as a function of time. When the velocity of a particle changes with time, the particle is said to be accelerating. For example, the magnitude of the velocity of a car increases when you step on the gas and decreases when you apply the brakes. Let us see how to quantify acceleration. Suppose an object that can be modeled as a particle moving along the x axis has an initial velocity vxi at time ti and a final velocity vxf at time tf , as in Figure 2.5a. – The average acceleration ax of the particle is defined as the change in velocity ⌬vx divided by the time interval ⌬t during which that change occurs: ax ϵ ⌬vx ⌬t ϭ vxf Ϫ vxi (2.6) tf Ϫ ti Average acceleration As with velocity, when the motion being analyzed is one-dimensional, we can use positive and negative signs to indicate the direction of the acceleration. Because the dimensions of velocity are L/T and the dimension of time is T, acceleration has dimensions of length divided by time squared, or L/T2. The SI unit of acceleration is meters per second squared (m/s2). It might be easier to interpret these units if you think of them as meters per second per second. For example, suppose an object has an acceleration of ϩ 2 m/s2. You should form a mental image of the object having a velocity that is along a straight line and is increasing by 2 m/s during every interval of 1 s. If the object starts from rest, you should be able to picture it moving at a velocity of ϩ 2 m/s after 1 s, at ϩ 4 m/s after 2 s, and so on. In some situations, the value of the average acceleration may be different over different time intervals. It is therefore useful to define the instantaneous acceleration as the limit of the average acceleration as ⌬t approaches zero. This concept is analogous to the definition of instantaneous velocity discussed in the previous section. If we imagine that point Ꭽ is brought closer and closer to point Ꭾ in Figure 2.5a and we take the limit of ⌬vx/⌬t as ⌬t approaches zero, we obtain the instantaneous acceleration: a x ϵ lim ⌬t : 0 ⌬vx dvx ϭ ⌬t dt Ꭾ vxf Ꭾ Ꭽ x tf v = vxf ti v = vxi (a) Instantaneous acceleration – ∆v ax = x ∆t vx vxi (2.7) ∆vx Ꭽ ∆t ti tf (b) t Figure 2.5 (a) A car, modeled as a particle, moving along the x axis from Ꭽ to Ꭾ has velocity vxi at t ϭ ti and velocity vxf at t ϭ tf . (b) Velocity–time graph (rust) for the particle moving in a straight line. The slope of the blue straight line connecting Ꭽ and Ꭾ is the average acceleration in the time interval ⌬t ϭ tf Ϫ ti.
    • C H A P T E R 2 • Motion in One Dimension 32 L PITFALL PREVENTION 2.4 Negative Acceleration Keep in mind that negative acceleration does not necessarily mean that an object is slowing down. If the acceleration is negative, and the velocity is negative, the object is speeding up! L PITFALL PREVENTION 2.5 Deceleration The word deceleration has the common popular connotation of slowing down. We will not use this word in this text, because it further confuses the definition we have given for negative acceleration. vx t tA tB That is, the instantaneous acceleration equals the derivative of the velocity with respect to time, which by definition is the slope of the velocity–time graph. The slope of the green line in Figure 2.5b is equal to the instantaneous acceleration at point Ꭾ. Thus, we see that just as the velocity of a moving particle is the slope at a point on the particle’s x -t graph, the acceleration of a particle is the slope at a point on the particle’s vx -t graph. One can interpret the derivative of the velocity with respect to time as the time rate of change of velocity. If ax is positive, the acceleration is in the positive x direction; if ax is negative, the acceleration is in the negative x direction. For the case of motion in a straight line, the direction of the velocity of an object and the direction of its acceleration are related as follows. When the object’s velocity and acceleration are in the same direction, the object is speeding up. On the other hand, when the object’s velocity and acceleration are in opposite directions, the object is slowing down. To help with this discussion of the signs of velocity and acceleration, we can relate the acceleration of an object to the force exerted on the object. In Chapter 5 we formally establish that force is proportional to acceleration: Fϰa This proportionality indicates that acceleration is caused by force. Furthermore, force and acceleration are both vectors and the vectors act in the same direction. Thus, let us think about the signs of velocity and acceleration by imagining a force applied to an object and causing it to accelerate. Let us assume that the velocity and acceleration are in the same direction. This situation corresponds to an object moving in some direction that experiences a force acting in the same direction. In this case, the object speeds up! Now suppose the velocity and acceleration are in opposite directions. In this situation, the object moves in some direction and experiences a force acting in the opposite direction. Thus, the object slows down! It is very useful to equate the direction of the acceleration to the direction of a force, because it is easier from our everyday experience to think about what effect a force will have on an object than to think only in terms of the direction of the acceleration. Quick Quiz 2.2 If a car is traveling eastward and slowing down, what is the direction of the force on the car that causes it to slow down? (a) eastward (b) westward (c) neither of these. tC (a) From now on we shall use the term acceleration to mean instantaneous acceleration. When we mean average acceleration, we shall always use the adjective average. Because vx ϭ dx/dt, the acceleration can also be written ax ax ϭ tC tA tB t (b) Figure 2.6 The instantaneous acceleration can be obtained from the velocity–time graph (a). At each instant, the acceleration in the ax versus t graph (b) equals the slope of the line tangent to the vx versus t curve (a). dvx d ϭ dt dt ΂ dx ΃ ϭ ddt x dt 2 2 (2.8) That is, in one-dimensional motion, the acceleration equals the second derivative of x with respect to time. Figure 2.6 illustrates how an acceleration–time graph is related to a velocity–time graph. The acceleration at any time is the slope of the velocity–time graph at that time. Positive values of acceleration correspond to those points in Figure 2.6a where the velocity is increasing in the positive x direction. The acceleration reaches a maximum at time tA, when the slope of the velocity–time graph is a maximum. The acceleration then goes to zero at time t B, when the velocity is a maximum (that is, when the slope of the vx -t graph is zero). The acceleration is negative when the velocity is decreasing in the positive x direction, and it reaches its most negative value at time t C.
    • S E C T I O N 2 . 3 • Acceleration 33 Quick Quiz 2.3 Make a velocity–time graph for the car in Figure 2.1a. The speed limit posted on the road sign is 30 km/h. True or false? The car exceeds the speed limit at some time within the interval. Conceptual Example 2.4 Graphical Relationships between x, vx , and ax The position of an object moving along the x axis varies with time as in Figure 2.7a. Graph the velocity versus time and the acceleration versus time for the object. Solution The velocity at any instant is the slope of the tangent to the x -t graph at that instant. Between t ϭ 0 and t ϭ tA, the slope of the x -t graph increases uniformly, and so the velocity increases linearly, as shown in Figure 2.7b. Between tA and t B, the slope of the x -t graph is constant, and so the velocity remains constant. At t D, the slope of the x -t graph is zero, so the velocity is zero at that instant. Between t D and t E, the slope of the x -t graph and thus the velocity are negative and decrease uniformly in this interval. In the interval t E to t F, the slope of the x-t graph is still negative, and at t F it goes to zero. Finally, after t F, the slope of the x -t graph is zero, meaning that the object is at rest for t Ͼ t F . The acceleration at any instant is the slope of the tangent to the vx -t graph at that instant. The graph of acceleration versus time for this object is shown in Figure 2.7c. The acceleration is constant and positive between 0 and tA, where the slope of the vx -t graph is positive. It is zero between tA and t B and for t Ͼ t F because the slope of the vx -t graph is zero at these times. It is negative between t B and t E because the slope of the vx -t graph is negative during this interval. Note that the sudden changes in acceleration shown in Figure 2.7c are unphysical. Such instantaneous changes cannot occur in reality. Example 2.5 x (a) tA tB tC tD tE tF tA O tB tC tD tE tF t vx (b) O t ax (c) O tB tA tE tF t Figure 2.7 (Example 2.4) (a) Position–time graph for an object moving along the x axis. (b) The velocity–time graph for the object is obtained by measuring the slope of the position–time graph at each instant. (c) The acceleration–time graph for the object is obtained by measuring the slope of the velocity–time graph at each instant. Average and Instantaneous Acceleration The velocity of a particle moving along the x axis varies in time according to the expression vx ϭ (40 Ϫ 5t 2 ) m/s, where t is in seconds. (A) Find the average acceleration in the time interval t ϭ 0 to t ϭ 2.0 s. Solution Figure 2.8 is a vx -t graph that was created from the velocity versus time expression given in the problem statement. Because the slope of the entire vx -t curve is negative, we expect the acceleration to be negative. We find the velocities at ti ϭ tA ϭ 0 and tf ϭ t B ϭ 2.0 s by substituting these values of t into the expression for the velocity: vx A ϭ (40 Ϫ 5t A2) m/s ϭ [40 Ϫ 5(0)2] m/s ϭ ϩ 40 m/s vx B ϭ (40 Ϫ 5t B2) m/s ϭ [40 Ϫ 5(2.0)2] m/s ϭ ϩ 20 m/s Therefore, the average acceleration in the specified time interval ⌬t ϭ t B Ϫ t A ϭ 2.0 s is ax ϭ vxf Ϫ vxi tf Ϫ ti ϭ vx B Ϫ vx A t B Ϫ tA ϭ (20 Ϫ 40) m/s (2.0 Ϫ 0) s ϭ Ϫ10 m/s2 The negative sign is consistent with our expectations— namely, that the average acceleration, which is represented by the slope of the line joining the initial and final points on the velocity–time graph, is negative. (B) Determine the acceleration at t ϭ 2.0 s.
    • 34 C H A P T E R 2 • Motion in One Dimension Solution The velocity at any time t is vxi ϭ (40 Ϫ 5t 2 ) m/s and the velocity at any later time t ϩ ⌬t is vx(m/s) 40 Ꭽ vxf ϭ 40 Ϫ 5(t ϩ ⌬t)2 ϭ 40 Ϫ 5t 2 Ϫ 10t ⌬t Ϫ 5(⌬t)2 30 Therefore, the change in velocity over the time interval ⌬t is ⌬vx ϭ vxf Ϫ vxi ϭ [Ϫ10t ⌬t Ϫ 5(⌬t)2] m/s Dividing this expression by ⌬t and taking the limit of the result as ⌬t approaches zero gives the acceleration at any time t: ⌬vx a x ϭ lim ϭ lim (Ϫ10t Ϫ 5⌬t) ϭ Ϫ10t m/s 2 ⌬t : 0 ⌬t ⌬t : 0 Therefore, at t ϭ 2.0 s, a x ϭ (Ϫ10)(2.0) m/s2 ϭ Ϫ20 m/s2 Because the velocity of the particle is positive and the acceleration is negative, the particle is slowing down. Note that the answers to parts (A) and (B) are different. The average acceleration in (A) is the slope of the blue line in Figure 2.8 connecting points Ꭽ and Ꭾ. The instantaneous acceleration in (B) is the slope of the green line tangent to the curve at point Ꭾ. Note also that the acceleration is not constant in this example. Situations involving constant acceleration are treated in Section 2.5. Slope = –20 m/s2 20 Ꭾ 10 t(s) 0 –10 –20 –30 0 1 2 3 4 Figure 2.8 (Example 2.5) The velocity–time graph for a particle moving along the x axis according to the expression vx ϭ (40 Ϫ 5t 2) m/s. The acceleration at t ϭ 2 s is equal to the slope of the green tangent line at that time. So far we have evaluated the derivatives of a function by starting with the definition of the function and then taking the limit of a specific ratio. If you are familiar with calculus, you should recognize that there are specific rules for taking derivatives. These rules, which are listed in Appendix B.6, enable us to evaluate derivatives quickly. For instance, one rule tells us that the derivative of any constant is zero. As another example, suppose x is proportional to some power of t, such as in the expression x ϭ At n where A and n are constants. (This is a very common functional form.) The derivative of x with respect to t is dx ϭ nAt nϪ1 dt Applying this rule to Example 2.5, in which vx ϭ 40 Ϫ 5t 2 , we find that the acceleration is ax ϭ dvx/dt ϭ Ϫ10t . 2.4 Motion Diagrams The concepts of velocity and acceleration are often confused with each other, but in fact they are quite different quantities. It is instructive to use motion diagrams to describe the velocity and acceleration while an object is in motion. A stroboscopic photograph of a moving object shows several images of the object, taken as the strobe light flashes at a constant rate. Figure 2.9 represents three sets of strobe photographs of cars moving along a straight roadway in a single direction, from left to right. The time intervals between flashes of the stroboscope are equal in each part of the diagram. In order not to confuse the two vector quantities, we use red for velocity vectors and violet for acceleration vectors in Figure 2.9. The vectors are
    • S E C T I O N 2 . 4 • Motion Diagrams 35 v (a) v (b) a v (c) a Active Figure 2.9 (a) Motion diagram for a car moving at constant velocity (zero acceleration). (b) Motion diagram for a car whose constant acceleration is in the direction of its velocity. The velocity vector at each instant is indicated by a red arrow, and the constant acceleration by a violet arrow. (c) Motion diagram for a car whose constant acceleration is in the direction opposite the velocity at each instant. sketched at several instants during the motion of the object. Let us describe the motion of the car in each diagram. In Figure 2.9a, the images of the car are equally spaced, showing us that the car moves through the same displacement in each time interval. This is consistent with the car moving with constant positive velocity and zero acceleration. We could model the car as a particle and describe it as a particle moving with constant velocity. In Figure 2.9b, the images become farther apart as time progresses. In this case, the velocity vector increases in time because the car’s displacement between adjacent positions increases in time. This suggests that the car is moving with a positive velocity and a positive acceleration. The velocity and acceleration are in the same direction. In terms of our earlier force discussion, imagine a force pulling on the car in the same direction it is moving—it speeds up. In Figure 2.9c, we can tell that the car slows as it moves to the right because its displacement between adjacent images decreases with time. In this case, this suggests that the car moves to the right with a constant negative acceleration. The velocity vector decreases in time and eventually reaches zero. From this diagram we see that the acceleration and velocity vectors are not in the same direction. The car is moving with a positive velocity but with a negative acceleration. (This type of motion is exhibited by a car that skids to a stop after applying its brakes.) The velocity and acceleration are in opposite directions. In terms of our earlier force discussion, imagine a force pulling on the car opposite to the direction it is moving—it slows down. The violet acceleration vectors in Figures 2.9b and 2.9c are all of the same length. Thus, these diagrams represent motion with constant acceleration. This is an important type of motion that will be discussed in the next section. Quick Quiz 2.4 Which of the following is true? (a) If a car is traveling eastward, its acceleration is eastward. (b) If a car is slowing down, its acceleration must be negative. (c) A particle with constant acceleration can never stop and stay stopped. At the Active Figures link at http://www.pse6.com, you can select the constant acceleration and initial velocity of the car and observe pictorial and graphical representations of its motion.
    • C H A P T E R 2 • Motion in One Dimension 36 2.5 One-Dimensional Motion with Constant Acceleration x Slope = vx f xi Slope = vxi t t 0 (a) If the acceleration of a particle varies in time, its motion can be complex and difficult to analyze. However, a very common and simple type of one-dimensional motion is that in which the acceleration is constant. When this is the case, the average acceleration ax over any time interval is numerically equal to the instantaneous acceleration ax at any instant within the interval, and the velocity changes at the same rate throughout the motion. If we replace ax by ax in Equation 2.6 and take ti ϭ 0 and tf to be any later time t, we find that vx ax ϭ Slope = ax or axt vx i vx i vx f t 0 vxf Ϫ vxi tϪ0 vxf ϭ vxi ϩ a xt ax Slope = 0 ax t (c) Active Figure 2.10 A particle moving along the x axis with constant acceleration ax; (a) the position–time graph, (b) the velocity–time graph, and (c) the acceleration–time graph. At the Active Figures link at http://www.pse6.com, you can adjust the constant acceleration and observe the effect on the position and velocity graphs. Position as a function of velocity and time (2.9) t (b) 0 (for constant a x) This powerful expression enables us to determine an object’s velocity at any time t if we know the object’s initial velocity vxi and its (constant) acceleration ax. A velocity–time graph for this constant-acceleration motion is shown in Figure 2.10b. The graph is a straight line, the (constant) slope of which is the acceleration ax ; this is consistent with the fact that ax ϭ dvx/dt is a constant. Note that the slope is positive; this indicates a positive acceleration. If the acceleration were negative, then the slope of the line in Figure 2.10b would be negative. When the acceleration is constant, the graph of acceleration versus time (Fig. 2.10c) is a straight line having a slope of zero. Because velocity at constant acceleration varies linearly in time according to Equation 2.9, we can express the average velocity in any time interval as the arithmetic mean of the initial velocity vxi and the final velocity vxf : vx ϭ vxi ϩ vxf 2 (for constant a x) (2.10) Note that this expression for average velocity applies only in situations in which the acceleration is constant. We can now use Equations 2.1, 2.2, and 2.10 to obtain the position of an object as a function of time. Recalling that ⌬x in Equation 2.2 represents xf Ϫ x i , and recognizing that ⌬t ϭ tf Ϫ ti ϭ t Ϫ 0 ϭ t, we find x f Ϫ x i ϭ vt ϭ 1 (vxi ϩ vxf )t 2 x f ϭ x i ϩ 1 (vxi ϩ vxf )t 2 (for constant a x) (2.11) This equation provides the final position of the particle at time t in terms of the initial and final velocities. We can obtain another useful expression for the position of a particle moving with constant acceleration by substituting Equation 2.9 into Equation 2.11: xf ϭ xi ϩ 1 [vxi ϩ (vxi ϩ axt )]t 2 Position as a function of time x f ϭ x i ϩ vxi t ϩ 1 a xt 2 2 (for constant a x ) (2.12) This equation provides the final position of the particle at time t in terms of the initial velocity and the acceleration. The position–time graph for motion at constant (positive) acceleration shown in Figure 2.10a is obtained from Equation 2.12. Note that the curve is a parabola.
    • S E C T I O N 2 . 5 • One-Dimensional Motion with Constant Acceleration The slope of the tangent line to this curve at t ϭ 0 equals the initial velocity vxi , and the slope of the tangent line at any later time t equals the velocity vxf at that time. Finally, we can obtain an expression for the final velocity that does not contain time as a variable by substituting the value of t from Equation 2.9 into Equation 2.11: x f ϭ x i ϩ 1 (vxi ϩ vxf ) 2 ΂ vxf Ϫ vxi v xf 2 ϭ v xi 2 ϩ 2a x (x f Ϫ x i ) ax ΃ ϭ v xf 2 Ϫ v xi 2 2a x (for constant a x) (2.13) This equation provides the final velocity in terms of the acceleration and the displacement of the particle. For motion at zero acceleration, we see from Equations 2.9 and 2.12 that vxf ϭ vxi ϭ vx xf ϭ xi ϩ vxt } when ax ϭ 0 That is, when the acceleration of a particle is zero, its velocity is constant and its position changes linearly with time. Quick Quiz 2.5 In Figure 2.11, match each vx -t graph on the left with the ax -t graph on the right that best describes the motion. ax vx t t (a) (d) vx ax t t (b) vx Active Figure 2.11 (Quick Quiz 2.5) Parts (a), (b), and (c) are vx -t graphs of objects in onedimensional motion. The possible accelerations of each object as a function of time are shown in scrambled order in (d), (e), and (f). (e) ax t (c) t (f ) At the Active Figures link at http://www.pse6.com, you can practice matching appropriate velocity vs. time graphs and acceleration vs. time graphs. Equations 2.9 through 2.13 are kinematic equations that may be used to solve any problem involving one-dimensional motion at constant acceleration. Keep in mind that these relationships were derived from the definitions of velocity and Velocity as a function of position 37
    • 38 C H A P T E R 2 • Motion in One Dimension Table 2.2 Kinematic Equations for Motion of a Particle Under Constant Acceleration Equation Information Given by Equation vxf ϭ vxi ϩ a xt x f ϭ x i ϩ 1(vxi ϩ vxf )t 2 x f ϭ x i ϩ vxi t ϩ 1ax t 2 2 v xf 2 ϭ v xi 2 ϩ 2a x(x f Ϫx i) Velocity as a function of time Position as a function of velocity and time Position as a function of time Velocity as a function of position Note: Motion is along the x axis. acceleration, together with some simple algebraic manipulations and the requirement that the acceleration be constant. The four kinematic equations used most often are listed in Table 2.2 for convenience. The choice of which equation you use in a given situation depends on what you know beforehand. Sometimes it is necessary to use two of these equations to solve for two unknowns. For example, suppose initial velocity vxi and acceleration ax are given. You can then find (1) the velocity at time t, using vxf ϭ vxi ϩ axt and (2) the position at time t, using xf ϭ xi ϩ vxit ϩ 1axt2. You should recognize that the quantities 2 that vary during the motion are position, velocity, and time. You will gain a great deal of experience in the use of these equations by solving a number of exercises and problems. Many times you will discover that more than one method can be used to obtain a solution. Remember that these equations of kinematics cannot be used in a situation in which the acceleration varies with time. They can be used only when the acceleration is constant. Example 2.6 Entering the Traffic Flow (A) Estimate your average acceleration as you drive up the entrance ramp to an interstate highway. Solution This problem involves more than our usual amount of estimating! We are trying to come up with a value of ax , but that value is hard to guess directly. The other variables involved in kinematics are position, velocity, and time. Velocity is probably the easiest one to approximate. Let us assume a final velocity of 100 km/h, so that you can merge with traffic. We multiply this value by (1 000 m/1 km) to convert kilometers to meters and then multiply by (1 h/3 600 s) to convert hours to seconds. These two calculations together are roughly equivalent to dividing by 3. In fact, let us just say that the final velocity is vxf Ϸ 30 m/s. (Remember, this type of approximation and the dropping of digits when performing estimations is okay. If you were starting with U.S. customary units, you could approximate 1 mi/h as roughly 0.5 m/s and continue from there.) Now we assume that you started up the ramp at about one third your final velocity, so that vxi Ϸ 10 m/s. Finally, we assume that it takes about 10 s to accelerate from vxi to vxf , basing this guess on our previous experience in automobiles. We can then find the average acceleration, using Equation 2.6: ax ϭ vxf Ϫ vxi t ϭ 2 m/s2 Ϸ 30 m/s Ϫ 10 m/s 10 s Granted, we made many approximations along the way, but this type of mental effort can be surprisingly useful and often yields results that are not too different from those derived from careful measurements. Do not be afraid to attempt making educated guesses and doing some fairly drastic number rounding to simplify estimations. Physicists engage in this type of thought analysis all the time. (B) How far did you go during the first half of the time interval during which you accelerated? Solution Let us assume that the acceleration is constant, with the value calculated in part (A). Because the motion takes place in a straight line and the velocity is always in the same direction, the distance traveled from the starting point is equal to the final position of the car. We can calculate the final position at 5 s from Equation 2.12: xf ϭ xi ϩ vxit ϩ 1 axt2 2 Ϸ 0 ϩ (10 m/s)(5 s) ϩ 1 (2 m/s2)(5 s)2 ϭ 50 m ϩ 25 m 2 ϭ 75 m This result indicates that if you had not accelerated, your initial velocity of 10 m/s would have resulted in a 50-m movement up the ramp during the first 5 s. The additional 25 m is the result of your increasing velocity during that interval.
    • S E C T I O N 2 . 5 • One-Dimensional Motion with Constant Acceleration Example 2.7 Carrier Landing A jet lands on an aircraft carrier at 140 mi/h (Ϸ 63 m/s). (A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the airplane and brings it to a stop? Solution We define our x axis as the direction of motion of the jet. A careful reading of the problem reveals that in addition to being given the initial speed of 63 m/s, we also know that the final speed is zero. We also note that we have no information about the change in position of the jet while it is slowing down. Equation 2.9 is the only equation in Table 2.2 that does not involve position, and so we use it to find the acceleration of the jet, modeled as a particle: ax ϭ vxf Ϫ vxi t Ϸ 0 Ϫ 63 m/s 2.0 s ϭ Ϫ31 m/s2 (B) If the plane touches down at position xi ϭ 0, what is the final position of the plane? Solution We can now use any of the other three equations in Table 2.2 to solve for the final position. Let us choose Equation 2.11: If the plane travels much farther than this, it might fall into the ocean. The idea of using arresting cables to slow down landing aircraft and enable them to land safely on ships originated at about the time of the first World War. The cables are still a vital part of the operation of modern aircraft carriers. What If? Suppose the plane lands on the deck of the aircraft carrier with a speed higher than 63 m/s but with the same acceleration as that calculated in part (A). How will that change the answer to part (B)? Answer If the plane is traveling faster at the beginning, it will stop farther away from its starting point, so the answer to part (B) should be larger. Mathematically, we see in Equation 2.11 that if vxi is larger, then xf will be larger. If the landing deck has a length of 75 m, we can find the maximum initial speed with which the plane can land and still come to rest on the deck at the given acceleration from Equation 2.13: v xf 2 ϭ v xi 2 ϩ 2a x (x f Ϫ x i) : vxi ϭ √vxf2 Ϫ 2a x (x f Ϫ x i) ϭ √0 Ϫ 2(Ϫ31 m/s2)(75 m Ϫ 0) x f ϭ x i ϩ 1(vxi ϩ vxf )t ϭ 0 ϩ 1(63 m/s ϩ 0)(2.0 s) 2 2 ϭ 68 m/s ϭ 63 m Example 2.8 Interactive Watch Out for the Speed Limit! A car traveling at a constant speed of 45.0 m/s passes a trooper hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch it, accelerating at a constant rate of 3.00 m/s2. How long does it take her to overtake the car? Solution Let us model the car and the trooper as particles. A sketch (Fig. 2.12) helps clarify the sequence of events. First, we write expressions for the position of each vehicle as a function of time. It is convenient to choose the position of the billboard as the origin and to set t B ϭ 0 as the time the trooper begins moving. At that instant, the car has already traveled a distance of 45.0 m because it has traveled at a constant speed of vx ϭ 45.0 m/s for 1 s. Thus, the initial position of the speeding car is x B ϭ 45.0 m. Because the car moves with constant speed, its acceleration is zero. Applying Equation 2.12 (with ax ϭ 0) gives for the car’s position at any time t: time t can be found from Equation 2.12: xf ϭ xi ϩ vxit ϩ 1 ax t 2 2 xtrooper ϭ 0 ϩ (0)t ϩ 1 axt 2 ϭ 1 (3.00 m/s2)t 2 2 2 v x car = 45.0 m/s a x car = 0 ax trooper = 3.00 m/s 2 tA = –1.00 s tB = 0 tC = ? Ꭽ Ꭾ Ꭿ x car ϭ x B ϩ vx cart ϭ 45.0 m ϩ (45.0 m/s)t A quick check shows that at t ϭ 0, this expression gives the car’s correct initial position when the trooper begins to move: x car ϭ x B ϭ 45.0 m. The trooper starts from rest at t B ϭ 0 and accelerates at 3.00 m/s2 away from the origin. Hence, her position at any Figure 2.12 (Example 2.8) A speeding car passes a hidden trooper. 39
    • C H A P T E R 2 • Motion in One Dimension 40 The trooper overtakes the car at the instant her position matches that of the car, which is position Ꭿ: x trooper ϭ x car 1 (3.00 2 time will be less than 31 s. Mathematically, let us cast the final quadratic equation above in terms of the parameters in the problem: m/s2)t 2 ϭ 45.0 m ϩ (45.0 m/s)t This gives the quadratic equation 1.50t 2 Ϫ 45.0t Ϫ 45.0 ϭ 0 The positive solution of this equation is t ϭ 31.0 s. 1 a t2 2 x The solution to this quadratic equation is, tϭ (For help in solving quadratic equations, see Appendix B.2.) ϭ What If? What if the trooper had a more powerful motorcycle with a larger acceleration? How would that change the time at which the trooper catches the car? Answer If the motorcycle has a larger acceleration, the trooper will catch up to the car sooner, so the answer for the Ϫ vx cart Ϫ x B ϭ 0 vx car Ϯ √v 2 car ϩ 2a x x B x ax vx car ϩ ax √ v 2 car 2x B x ϩ ax 2 ax where we have chosen the positive sign because that is the only choice consistent with a time t Ͼ 0. Because all terms on the right side of the equation have the acceleration ax in the denominator, increasing the acceleration will decrease the time at which the trooper catches the car. You can study the motion of the car and trooper for various velocities of the car at the Interactive Worked Example link at http://www.pse6.com. 2.6 Freely Falling Objects Galileo Galilei Italian physicist and astronomer (1564–1642) Galileo formulated the laws that govern the motion of objects in free fall and made many other significant discoveries in physics and astronomy. Galileo publicly defended Nicholaus Copernicus’s assertion that the Sun is at the center of the Universe (the heliocentric system). He published Dialogue Concerning Two New World Systems to support the Copernican model, a view which the Church declared to be heretical. (North Wind) L PITFALL PREVENTION 2.6 g and g Be sure not to confuse the italicized symbol g for free-fall acceleration with the nonitalicized symbol g used as the abbreviation for “gram.” It is well known that, in the absence of air resistance, all objects dropped near the Earth’s surface fall toward the Earth with the same constant acceleration under the influence of the Earth’s gravity. It was not until about 1600 that this conclusion was accepted. Before that time, the teachings of the great philosopher Aristotle (384–322 B.C.) had held that heavier objects fall faster than lighter ones. The Italian Galileo Galilei (1564–1642) originated our present-day ideas concerning falling objects. There is a legend that he demonstrated the behavior of falling objects by observing that two different weights dropped simultaneously from the Leaning Tower of Pisa hit the ground at approximately the same time. Although there is some doubt that he carried out this particular experiment, it is well established that Galileo performed many experiments on objects moving on inclined planes. In his experiments he rolled balls down a slight incline and measured the distances they covered in successive time intervals. The purpose of the incline was to reduce the acceleration; with the acceleration reduced, Galileo was able to make accurate measurements of the time intervals. By gradually increasing the slope of the incline, he was finally able to draw conclusions about freely falling objects because a freely falling ball is equivalent to a ball moving down a vertical incline. You might want to try the following experiment. Simultaneously drop a coin and a crumpled-up piece of paper from the same height. If the effects of air resistance are negligible, both will have the same motion and will hit the floor at the same time. In the idealized case, in which air resistance is absent, such motion is referred to as free-fall. If this same experiment could be conducted in a vacuum, in which air resistance is truly negligible, the paper and coin would fall with the same acceleration even when the paper is not crumpled. On August 2, 1971, such a demonstration was conducted on the Moon by astronaut David Scott. He simultaneously released a hammer and a feather, and they fell together to the lunar surface. This demonstration surely would have pleased Galileo! When we use the expression freely falling object, we do not necessarily refer to an object dropped from rest. A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward or downward and those released from rest are all falling freely once they
    • S E C T I O N 2 . 6 • Freely Falling Objects are released. Any freely falling object experiences an acceleration directed downward, regardless of its initial motion. We shall denote the magnitude of the free-fall acceleration by the symbol g. The value of g near the Earth’s surface decreases with increasing altitude. Furthermore, slight variations in g occur with changes in latitude. It is common to define “up” as the ϩ y direction and to use y as the position variable in the kinematic equations. At the Earth’s surface, the value of g is approximately 9.80 m/s2. Unless stated otherwise, we shall use this value for g when performing calculations. For making quick estimates, use g ϭ 10 m/s2. If we neglect air resistance and assume that the free-fall acceleration does not vary with altitude over short vertical distances, then the motion of a freely falling object moving vertically is equivalent to motion in one dimension under constant acceleration. Therefore, the equations developed in Section 2.5 for objects moving with constant acceleration can be applied. The only modification that we need to make in these equations for freely falling objects is to note that the motion is in the vertical direction (the y direction) rather than in the horizontal direction (x) and that the acceleration is downward and has a magnitude of 9.80 m/s2. Thus, we always choose ay ϭ Ϫ g ϭ Ϫ 9.80 m/s2, where the negative sign means that the acceleration of a freely falling object is downward. In Chapter 13 we shall study how to deal with variations in g with altitude. Quick Quiz 2.6 A ball is thrown upward. While the ball is in free fall, does its acceleration (a) increase (b) decrease (c) increase and then decrease (d) decrease and then increase (e) remain constant? Quick Quiz 2.7 After a ball is thrown upward and is in the air, its speed (a) increases (b) decreases (c) increases and then decreases (d) decreases and then increases (e) remains the same. Conceptual Example 2.9 PITFALL PREVENTION 2.7 The Sign of g Keep in mind that g is a positive number—it is tempting to substitute Ϫ 9.80 m/s2 for g, but resist the temptation. Downward gravitational acceleration is indicated explicitly by stating the acceleration as ay ϭ Ϫ g. L PITFALL PREVENTION 2.8 Acceleration at the Top of The Motion It is a common misconception that the acceleration of a projectile at the top of its trajectory is zero. While the velocity at the top of the motion of an object thrown upward momentarily goes to zero, the acceleration is still that due to gravity at this point. If the velocity and acceleration were both zero, the projectile would stay at the top! The Daring Sky Divers A sky diver jumps out of a hovering helicopter. A few seconds later, another sky diver jumps out, and they both fall along the same vertical line. Ignore air resistance, so that both sky divers fall with the same acceleration. Does the difference in their speeds stay the same throughout the fall? Does the vertical distance between them stay the same throughout the fall? Solution At any given instant, the speeds of the divers are different because one had a head start. In any time interval Example 2.10 L 41 ⌬t after this instant, however, the two divers increase their speeds by the same amount because they have the same acceleration. Thus, the difference in their speeds remains the same throughout the fall. The first jumper always has a greater speed than the second. Thus, in a given time interval, the first diver covers a greater distance than the second. Consequently, the separation distance between them increases. Describing the Motion of a Tossed Ball A ball is tossed straight up at 25 m/s. Estimate its velocity at 1-s intervals. Solution Let us choose the upward direction to be positive. Regardless of whether the ball is moving upward or downward, its vertical velocity changes by approximately Ϫ 10 m/s for every second it remains in the air. It starts out at 25 m/s. After 1 s has elapsed, it is still moving upward but at 15 m/s because its acceleration is downward (downward acceleration causes its velocity to decrease). After another second, its upward velocity has dropped to 5 m/s. Now comes the tricky part—after another half second, its velocity is zero. The ball has gone as high as it will go. After the last half of this 1-s interval, the ball is moving at Ϫ 5 m/s. (The negative sign tells us that the ball is now moving in the negative direction, that is, downward. Its velocity has changed from ϩ 5 m/s to Ϫ 5 m/s during that 1-s interval. The change in velocity is still Ϫ 5 m/s Ϫ (ϩ 5 m/s) ϭ Ϫ 10 m/s in that second.) It continues downward, and after another 1 s has elapsed, it is falling at a velocity of Ϫ 15 m/s. Finally, after another 1 s, it has reached its original starting point and is moving downward at Ϫ 25 m/s.
    • 42 C H A P T E R 2 • Motion in One Dimension Conceptual Example 2.11 Follow the Bouncing Ball A tennis ball is dropped from shoulder height (about 1.5 m) and bounces three times before it is caught. Sketch graphs of its position, velocity, and acceleration as functions of time, with the ϩ y direction defined as upward. interval, and so the acceleration must be quite large and positive. This corresponds to the very steep upward lines on the velocity–time graph and to the spikes on the acceleration–time graph. Solution For our sketch let us stretch things out horizontally so that we can see what is going on. (Even if the ball were moving horizontally, this motion would not affect its vertical motion.) From Figure 2.13a we see that the ball is in contact with the floor at points Ꭾ, ൳, and ൵. Because the velocity of the ball changes from negative to positive three times during these bounces (Fig. 2.13b), the slope of the position–time graph must change in the same way. Note that the time interval between bounces decreases. Why is that? During the rest of the ball’s motion, the slope of the velocity–time graph in Fig. 2.13b should be Ϫ 9.80 m/s2 . The acceleration–time graph is a horizontal line at these times because the acceleration does not change when the ball is in free fall. When the ball is in contact with the floor, the velocity changes substantially during a very short time tA tB tC tD tE tF y(m) 1 t(s) 0 vy (m/s) 4 t(s) 0 –4 1.5 Ꭽ t(s) Ꭿ 1.0 ay (m/s2) –4 ൴ 0.5 –8 0.0 Ꭾ ൳ ൵ –12 (b) (a) Active Figure 2.13 (Conceptual Example 2.11) (a) A ball is dropped from a height of 1.5 m and bounces from the floor. (The horizontal motion is not considered here because it does not affect the vertical motion.) (b) Graphs of position, velocity, and acceleration versus time. At the Active Figures link at http://www.pse6.com, you can adjust both the value for g and the amount of “bounce” of the ball, and observe the resulting motion of the ball both pictorially and graphically. Quick Quiz 2.8 Which values represent the ball’s vertical velocity and acceleration at points Ꭽ, Ꭿ, and ൴ in Figure 2.13a? (a) vy ϭ 0, ay ϭ Ϫ 9.80 m/s2 (b) vy ϭ 0, ay ϭ 9.80 m/s2 (c) vy ϭ 0, ay ϭ 0 (d) vy ϭ Ϫ 9.80 m/s, ay ϭ 0
    • S E C T I O N 2 . 6 • Freely Falling Objects Example 2.12 43 Interactive Not a Bad Throw for a Rookie! A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The building is 50.0 m high, and the stone just misses the edge of the roof on its way down, as shown in Figure 2.14. Using tA ϭ 0 as the time the stone leaves the thrower’s hand at position Ꭽ, determine (A) the time at which the stone reaches its maximum height, (B) the maximum height, (C) the time at which the stone returns to the height from which it was thrown, (D) the velocity of the stone at this instant, and (E) the velocity and position of the stone at t ϭ 5.00 s. Ꭾ tA = 0 yA = 0 vy A = 20.0 m/s ay A = –9.80 m/s2 tB = 2.04 s yB = 20.4 m vy B = 0 ay B = –9.80 m/s2 Ꭽ Ꭿ tC = 4.08 s yC = 0 vy C = –20.0 m/s ay C = –9.80 m/s2 ൳ tD = 5.00 s yD = –22.5 m vy D = –29.0 m/s ay D = –9.80 m/s2 Ꭽ Solution (A) As the stone travels from Ꭽ to Ꭾ, its velocity must change by 20 m/s because it stops at Ꭾ. Because gravity causes vertical velocities to change by about 10 m/s for every second of free fall, it should take the stone about 2 s to go from Ꭽ to Ꭾ in our drawing. To calculate the exact time tB at which the stone reaches maximum height, we use Equation 2.9, vy B ϭ vyA ϩ a yt , noting that vy B ϭ 0 and setting the start of our clock readings at tA ϭ 0: 0 ϭ 20.0 m/s ϩ (Ϫ 9.80 m/s2)t t ϭ tB ϭ 20.0 m/s ϭ 2.04 s 9.80 m/s2 50.0 m Our estimate was pretty close. (B) Because the average velocity for this first interval is 10 m/s (the average of 20 m/s and 0 m/s) and because it travels for about 2 s, we expect the stone to travel about 20 m. By substituting our time into Equation 2.12, we can find the maximum height as measured from the position of the thrower, where we set yA ϭ 0: ymax ϭ y B ϭ yA ϩ vx At ϩ 1 a yt 2 2 y B ϭ 0 ϩ (20.0 m/s)(2.04 s) ϩ 1(Ϫ9.80 m/s2)(2.04 s)2 2 ϭ 20.4 m ൴ Our free-fall estimates are very accurate. (C) There is no reason to believe that the stone’s motion from Ꭾ to Ꭿ is anything other than the reverse of its motion from Ꭽ to Ꭾ. The motion from Ꭽ to Ꭿ is symmetric. Thus, the time needed for it to go from Ꭽ to Ꭿ should be twice the time needed for it to go from Ꭽ to Ꭾ. When the stone is back at the height from which it was thrown (position Ꭿ), the y coordinate is again zero. Using Equation 2.12, with y C ϭ 0, we obtain y C ϭ y A ϩ vy At ϩ 1 a yt 2 2 0 ϭ 0 ϩ 20.0t Ϫ 4.90t 2 This is a quadratic equation and so has two solutions for t ϭ t C. The equation can be factored to give t(20.0 Ϫ 4.90t) ϭ 0 One solution is t ϭ 0, corresponding to the time the stone starts its motion. The other solution is t ϭ 4.08 s, which tE = 5.83 s yE = –50.0 m vy E = –37.1 m/s 2 ay E = –9.80 m/s Figure 2.14 (Example 2.12) Position and velocity versus time for a freely falling stone thrown initially upward with a velocity vyi ϭ 20.0 m/s. is the solution we are after. Notice that it is double the value we calculated for t B. (D) Again, we expect everything at Ꭿ to be the same as it is at Ꭽ, except that the velocity is now in the opposite direction. The value for t found in (c) can be inserted into Equation 2.9 to give vy C ϭ vy A ϩ a y t ϭ 20.0 m/s ϩ (Ϫ9.80 m/s2)(4.08 s) ϭ Ϫ20.0 m/s The velocity of the stone when it arrives back at its original height is equal in magnitude to its initial velocity but opposite in direction.
    • 44 C H A P T E R 2 • Motion in One Dimension (E) For this part we ignore the first part of the motion (Ꭽ : Ꭾ) and consider what happens as the stone falls from position Ꭾ, where it has zero vertical velocity, to position ൳. We define the initial time as t B ϭ 0. Because the given time for this part of the motion relative to our new zero of time is 5.00 s Ϫ 2.04 s ϭ 2.96 s, we estimate that the acceleration due to gravity will have changed the speed by about 30 m/s. We can calculate this from Equation 2.9, where we take t ϭ 2.96 s: vy D ϭ vy B ϩ a y t ϭ 0 m/s ϩ (Ϫ 9.80 m/s2)(2.96 s) position of the stone at t D ϭ 5.00 s (with respect to t A ϭ 0) by defining a new initial instant, t C ϭ 0: y D ϭ y C ϩ vy C t ϩ 1 a y t 2 2 ϭ 0 ϩ (Ϫ20.0 m/s)(5.00 s Ϫ 4.08 s) ϩ 1 (Ϫ 9.80 m/s2)(5.00 s Ϫ4.08 s)2 2 ϭ Ϫ22.5 m What If? What if the building were 30.0 m tall instead of 50.0 m tall? Which answers in parts (A) to (E) would change? ϭ Ϫ29.0 m/s We could just as easily have made our calculation between positions Ꭽ (where we return to our original initial time tA ϭ 0) and ൳: vy D ϭ vy A ϩ a y t ϭ 20.0 m/s ϩ (Ϫ 9.80 m/s2)(5.00 s) ϭ Ϫ29.0 m/s To further demonstrate that we can choose different initial instants of time, let us use Equation 2.12 to find the Answer None of the answers would change. All of the motion takes place in the air, and the stone does not interact with the ground during the first 5.00 s. (Notice that even for a 30.0-m tall building, the stone is above the ground at t ϭ 5.00 s.) Thus, the height of the building is not an issue. Mathematically, if we look back over our calculations, we see that we never entered the height of the building into any equation. You can study the motion of the thrown ball at the Interactive Worked Example link at http://www.pse6.com. 2.7 Kinematic Equations Derived from Calculus This section assumes the reader is familiar with the techniques of integral calculus. If you have not yet studied integration in your calculus course, you should skip this section or cover it after you become familiar with integration. The velocity of a particle moving in a straight line can be obtained if its position as a function of time is known. Mathematically, the velocity equals the derivative of the position with respect to time. It is also possible to find the position of a particle if its velocity is known as a function of time. In calculus, the procedure used to perform this task is referred to either as integration or as finding the antiderivative. Graphically, it is equivalent to finding the area under a curve. Suppose the vx -t graph for a particle moving along the x axis is as shown in Figure 2.15. Let us divide the time interval tf Ϫ t i into many small intervals, each of vx Area = vxn ∆tn vxn ti tf ∆t n t Figure 2.15 Velocity versus time for a particle moving along the x axis. The area of the shaded rectangle is equal to the displacement ⌬x in the time interval ⌬tn , while the total area under the curve is the total displacement of the particle.
    • S E C T I O N 2 . 7 • Kinematic Equations Derived from Calculus duration ⌬tn. From the definition of average velocity we see that the displacement during any small interval, such as the one shaded in Figure 2.15, is given by ⌬x n ϭ vxn ⌬t n where vxn is the average velocity in that interval. Therefore, the displacement during this small interval is simply the area of the shaded rectangle. The total displacement for the interval tf Ϫ ti is the sum of the areas of all the rectangles: ⌬x ϭ ͚vxn ⌬t n n where the symbol ⌺ (upper case Greek sigma) signifies a sum over all terms, that is, over all values of n. In this case, the sum is taken over all the rectangles from ti to tf . Now, as the intervals are made smaller and smaller, the number of terms in the sum increases and the sum approaches a value equal to the area under the velocity–time graph. Therefore, in the limit n : ϱ, or ⌬tn : 0, the displacement is ͚vxn ⌬t n ⌬x ϭ lim ⌬tn : 0 n (2.14) or Displacement ϭ area under the vx-t graph Note that we have replaced the average velocity vxn with the instantaneous velocity vxn in the sum. As you can see from Figure 2.15, this approximation is valid in the limit of very small intervals. Therefore if we know the vx -t graph for motion along a straight line, we can obtain the displacement during any time interval by measuring the area under the curve corresponding to that time interval. The limit of the sum shown in Equation 2.14 is called a definite integral and is written ͚vxn ⌬t n ϭ :0 n lim ⌬tn ͵ tf ti vx(t)dt (2.15) where vx(t) denotes the velocity at any time t. If the explicit functional form of vx(t) is known and the limits are given, then the integral can be evaluated. Sometimes the vx -t graph for a moving particle has a shape much simpler than that shown in Figure 2.15. For example, suppose a particle moves at a constant velocity vxi. In this case, the vx -t graph is a horizontal line, as in Figure 2.16, and the displacement of the particle during the time interval ⌬t is simply the area of the shaded rectangle: ⌬x ϭ vxi ⌬t (when vx ϭ vxi ϭ constant) As another example, consider a particle moving with a velocity that is proportional to t, as in Figure 2.17. Taking vx ϭ axt, where ax is the constant of proportionality (the vx vx = vxi = constant ∆t vxi vxi ti tf t Figure 2.16 The velocity–time curve for a particle moving with constant velocity vxi. The displacement of the particle during the time interval tf Ϫ ti is equal to the area of the shaded rectangle. Definite integral 45
    • 46 C H A P T E R 2 • Motion in One Dimension vx Ꭽ v x = a xt a xtA Figure 2.17 The velocity–time curve for a particle moving with a velocity that is proportional to the time. t tA acceleration), we find that the displacement of the particle during the time interval t ϭ 0 to t ϭ tA is equal to the area of the shaded triangle in Figure 2.17: ⌬x ϭ 1(t A)(ax t A) ϭ 1 ax t A2 2 2 Kinematic Equations We now use the defining equations for acceleration and velocity to derive two of our kinematic equations, Equations 2.9 and 2.12. The defining equation for acceleration (Eq. 2.7), ax ϭ dvx dt may be written as dvx ϭ ax dt or, in terms of an integral (or antiderivative), as vxf Ϫ vxi ϭ ͵ t 0 a x dt For the special case in which the acceleration is constant, ax can be removed from the integral to give ͵ t vxf Ϫvxi ϭ a x dt ϭ a x(t Ϫ 0) ϭ a xt (2.16) 0 which is Equation 2.9. Now let us consider the defining equation for velocity (Eq. 2.5): vx ϭ dx dt We can write this as dx ϭ vx dt, or in integral form as x f Ϫx i ϭ ͵ t 0 vx dt Because vx ϭ vxf ϭ vxi ϩ axt, this expression becomes xf Ϫ xi ϭ ͵ t 0 (vxi ϩ a xt)dt ϭ ϭ vxi t ϩ 1 a t2 2 x ͵ t 0 ͵ t ΂ t2 vxi dt ϩ a x tdt ϭ vxi(t Ϫ 0) ϩ a x 0 2 ΃ Ϫ0 which is Equation 2.12. Besides what you might expect to learn about physics concepts, a very valuable skill you should hope to take away from your physics course is the ability to solve complicated problems. The way physicists approach complex situations and break them down into manageable pieces is extremely useful. On the next page is a general problem-solving strategy that will help guide you through the steps. To help you remember the steps of the strategy, they are called Conceptualize, Categorize, Analyze, and Finalize.
    • G E N E RAL P R O B L E M -S O LVI N G ST RAT E GY Analyze Conceptualize • The first thing to do when approaching a problem is to think about and understand the situation. Study carefully any diagrams, graphs, tables, or photographs that accompany the problem. Imagine a movie, running in your mind, of what happens in the problem. • If a diagram is not provided, you should almost always make a quick drawing of the situation. Indicate any known values, perhaps in a table or directly on your sketch. • • Now you must analyze the problem and strive for a mathematical solution. Because you have already categorized the problem, it should not be too difficult to select relevant equations that apply to the type of situation in the problem. For example, if the problem involves a particle moving under constant acceleration, Equations 2.9 to 2.13 are relevant. • Use algebra (and calculus, if necessary) to solve symbolically for the unknown variable in terms of what is given. Substitute in the appropriate numbers, calculate the result, and round it to the proper number of significant figures. Now focus on what algebraic or numerical information is given in the problem. Carefully read the problem statement, looking for key phrases such as “starts from rest” (vi ϭ 0), “stops” (vf ϭ 0), or “freely falls” (ay ϭ Ϫ g ϭ Ϫ 9.80 m/s2). • Now focus on the expected result of solving the problem. Exactly what is the question asking? Will the final result be numerical or algebraic? Do you know what units to expect? • Finalize Don’t forget to incorporate information from your own experiences and common sense. What should a reasonable answer look like? For example, you wouldn’t expect to calculate the speed of an automobile to be 5 ϫ 106 m/s. • This is the most important part. Examine your numerical answer. Does it have the correct units? Does it meet your expectations from your conceptualization of the problem? What about the algebraic form of the result—before you substituted numerical values? Does it make sense? Examine the variables in the problem to see whether the answer would change in a physically meaningful way if they were drastically increased or decreased or even became zero. Looking at limiting cases to see whether they yield expected values is a very useful way to make sure that you are obtaining reasonable results. • Think about how this problem compares with others you have solved. How was it similar? In what critical ways did it differ? Why was this problem assigned? You should have learned something by doing it. Can you figure out what? If it is a new category of problem, be sure you understand it so that you can use it as a model for solving future problems in the same category. Categorize • Once you have a good idea of what the problem is about, you need to simplify the problem. Remove the details that are not important to the solution. For example, model a moving object as a particle. If appropriate, ignore air resistance or friction between a sliding object and a surface. • Once the problem is simplified, it is important to categorize the problem. Is it a simple plug-in problem, such that numbers can be simply substituted into a definition? If so, the problem is likely to be finished when this substitution is done. If not, you face what we can call an analysis problem—the situation must be analyzed more deeply to reach a solution. • If it is an analysis problem, it needs to be categorized further. Have you seen this type of problem before? Does it fall into the growing list of types of problems that you have solved previously? Being able to classify a problem can make it much easier to lay out a plan to solve it. For example, if your simplification shows that the problem can be treated as a particle moving under constant acceleration and you have already solved such a problem (such as the examples in Section 2.5), the solution to the present problem follows a similar pattern. When solving complex problems, you may need to identify a series of sub-problems and apply the problemsolving strategy to each. For very simple problems, you probably don’t need this strategy at all. But when you are looking at a problem and you don’t know what to do next, remember the steps in the strategy and use them as a guide. For practice, it would be useful for you to go back over the examples in this chapter and identify the Conceptualize, Categorize, Analyze, and Finalize steps. In the next chapter, we will begin to show these steps explicitly in the examples. 47
    • 48 C H A P T E R 2 • Motion in One Dimension S U M MARY Take a practice test for this chapter by clicking the Practice Test link at http://www.pse6.com. After a particle moves along the x axis from some initial position xi to some final position xf , its displacement is ⌬x ϵ x f Ϫ x i (2.1) The average velocity of a particle during some time interval is the displacement ⌬x divided by the time interval ⌬t during which that displacement occurs: ⌬x ⌬t vx ϵ (2.2) The average speed of a particle is equal to the ratio of the total distance it travels to the total time interval during which it travels that distance: Average speed ϭ total distance total time (2.3) The instantaneous velocity of a particle is defined as the limit of the ratio ⌬x/⌬t as ⌬t approaches zero. By definition, this limit equals the derivative of x with respect to t, or the time rate of change of the position: vx ϵ lim ⌬t : 0 ⌬x dx ϭ ⌬t dt (2.5) The instantaneous speed of a particle is equal to the magnitude of its instantaneous velocity. The average acceleration of a particle is defined as the ratio of the change in its velocity ⌬vx divided by the time interval ⌬t during which that change occurs: ax ϵ vxf Ϫ vxi ⌬vx ϭ ⌬t tf Ϫ ti (2.6) The instantaneous acceleration is equal to the limit of the ratio ⌬vx /⌬t as ⌬t approaches 0. By definition, this limit equals the derivative of vx with respect to t, or the time rate of change of the velocity: a x ϵ lim ⌬t : 0 ⌬vx dvx ϭ ⌬t dt (2.7) When the object’s velocity and acceleration are in the same direction, the object is speeding up. On the other hand, when the object’s velocity and acceleration are in opposite directions, the object is slowing down. Remembering that F ϰ a is a useful way to identify the direction of the acceleration. The equations of kinematics for a particle moving along the x axis with uniform acceleration ax (constant in magnitude and direction) are vxf ϭ vxi ϩ axt xf ϭ xi ϩ vxt ϭ xi ϩ 1 (v 2 xi (2.9) ϩ vxf)t (2.11) x f ϭ x i ϩ vxi t ϩ 1a x t 2 2 (2.12) v xf 2 ϭ v xi 2 ϩ 2a x(x f Ϫ x i) (2.13) An object falling freely in the presence of the Earth’s gravity experiences a free-fall acceleration directed toward the center of the Earth. If air resistance is neglected, if the motion occurs near the surface of the Earth, and if the range of the motion is small compared with the Earth’s radius, then the free-fall acceleration g is constant over the range of motion, where g is equal to 9.80 m/s2. Complicated problems are best approached in an organized manner. You should be able to recall and apply the Conceptualize, Categorize, Analyze, and Finalize steps of the General Problem-Solving Strategy when you need them.
    • Problems 49 QU ESTIONS 1. The speed of sound in air is 331 m/s. During the next thunderstorm, try to estimate your distance from a lightning bolt by measuring the time lag between the flash and the thunderclap. You can ignore the time it takes for the light flash to reach you. Why? 2. The average velocity of a particle moving in one dimension has a positive value. Is it possible for the instantaneous velocity to have been negative at any time in the interval? Suppose the particle started at the origin x ϭ 0. If its average velocity is positive, could the particle ever have been in the Ϫ x region of the axis? 3. If the average velocity of an object is zero in some time interval, what can you say about the displacement of the object for that interval? 4. Can the instantaneous velocity of an object at an instant of time ever be greater in magnitude than the average velocity over a time interval containing the instant? Can it ever be less? 5. If an object’s average velocity is nonzero over some time interval, does this mean that its instantaneous velocity is never zero during the interval? Explain your answer. 6. If an object’s average velocity is zero over some time interval, show that its instantaneous velocity must be zero at some time during the interval. It may be useful in your proof to sketch a graph of x versus t and to note that vx(t ) is a continuous function. 7. If the velocity of a particle is nonzero, can its acceleration be zero? Explain. 8. If the velocity of a particle is zero, can its acceleration be nonzero? Explain. 9. Two cars are moving in the same direction in parallel lanes along a highway. At some instant, the velocity of car A exceeds the velocity of car B. Does this mean that the acceleration of A is greater than that of B? Explain. 10. Is it possible for the velocity and the acceleration of an object to have opposite signs? If not, state a proof. If so, give an example of such a situation and sketch a velocity–time graph to prove your point. 11. Consider the following combinations of signs and values for velocity and acceleration of a particle with respect to a one-dimensional x axis: Velocity a. b. c. d. e. f. g. h. Acceleration Positive Positive Positive Negative Negative Negative Zero Zero Positive Negative Zero Positive Negative Zero Positive Negative Describe what a particle is doing in each case, and give a real life example for an automobile on an east-west one-dimensional axis, with east considered the positive direction. 12. Can the equations of kinematics (Eqs. 2.9–2.13) be used in a situation where the acceleration varies in time? Can they be used when the acceleration is zero? 13. A stone is thrown vertically upward from the roof of a building. Does the position of the stone depend on the location chosen for the origin of the coordinate system? Does the stone’s velocity depend on the choice of origin? Explain your answers. 14. A child throws a marble into the air with an initial speed vi. Another child drops a ball at the same instant. Compare the accelerations of the two objects while they are in flight. 15. A student at the top of a building of height h throws one ball upward with a speed of vi and then throws a second ball downward with the same initial speed, vi . How do the final velocities of the balls compare when they reach the ground? 16. An object falls freely from height h. It is released at time zero and strikes the ground at time t. (a) When the object is at height 0.5h, is the time earlier than 0.5t, equal to 0.5t, or later than 0.5t? (b) When the time is 0.5t, is the height of the object greater than 0.5h, equal to 0.5h, or less than 0.5h? Give reasons for your answers. 17. You drop a ball from a window on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you have a friend down on the street who throws another ball upward at speed v. Your friend throws the ball upward at exactly the same time that you drop yours from the window. At some location, the balls pass each other. Is this location at the halfway point between window and ground, above this point, or below this point? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide = coached solution with hints available at http://www.pse6.com = computer useful in solving problem = paired numerical and symbolic problems Section 2.1 Position, Velocity, and Speed 1. The position of a pinewood derby car was observed at various times; the results are summarized in the following table. Find the average velocity of the car for (a) the first second, (b) the last 3 s, and (c) the entire period of observation. t(s) 0 1.0 2.0 3.0 4.0 5.0 x(m) 0 2.3 9.2 20.7 36.8 57.5
    • 50 C H A P T E R 2 • Motion in One Dimension 2. (a) Sand dunes in a desert move over time as sand is swept up the windward side to settle in the lee side. Such “walking” dunes have been known to walk 20 feet in a year and can travel as much as 100 feet per year in particularly windy times. Calculate the average speed in each case in m/s. (b) Fingernails grow at the rate of drifting continents, on the order of 10 mm/yr. Approximately how long did it take for North America to separate from Europe, a distance of about 3 000 mi? x(m) 12 10 8 6 4 3. The position versus time for a certain particle moving along the x axis is shown in Figure P2.3. Find the average velocity in the time intervals (a) 0 to 2 s, (b) 0 to 4 s, (c) 2 s to 4 s, (d) 4 s to 7 s, (e) 0 to 8 s. 2 0 1 x(m) 6 4 2 1 2 3 4 5 6 7 8 t(s) –2 –4 –6 Figure P2.3 Problems 3 and 9 5 6 t(s) 10. A hare and a tortoise compete in a race over a course 1.00 km long. The tortoise crawls straight and steadily at its maximum speed of 0.200 m/s toward the finish line. The hare runs at its maximum speed of 8.00 m/s toward the goal for 0.800 km and then stops to tease the tortoise. How close to the goal can the hare let the tortoise approach before resuming the race, which the tortoise wins in a photo finish? Assume that, when moving, both animals move steadily at their respective maximum speeds. Section 2.3 Acceleration 10t 2 4. A particle moves according to the equation x ϭ where x is in meters and t is in seconds. (a) Find the average velocity for the time interval from 2.00 s to 3.00 s. (b) Find the average velocity for the time interval from 2.00 to 2.10 s. 5. A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3.00 m/s. What is (a) her average speed over the entire trip? (b) her average velocity over the entire trip? Instantaneous Velocity and Speed 6. The position of a particle moving along the x axis varies in time according to the expression x ϭ 3t 2, where x is in meters and t is in seconds. Evaluate its position (a) at t ϭ 3.00 s and (b) at 3.00 s ϩ ⌬t. (c) Evaluate the limit of ⌬x/⌬t as ⌬t approaches zero, to find the velocity at t ϭ 3.00 s. 7. 4 9. Find the instantaneous velocity of the particle described in Figure P2.3 at the following times: (a) t ϭ 1.0 s, (b) t ϭ 3.0 s, (c) t ϭ 4.5 s, and (d) t ϭ 7.5 s. 8 Section 2.2 3 Figure P2.7 10 0 2 A position-time graph for a particle moving along the x axis is shown in Figure P2.7. (a) Find the average velocity in the time interval t ϭ 1.50 s to t ϭ 4.00 s. (b) Determine the instantaneous velocity at t ϭ 2.00 s by measuring the slope of the tangent line shown in the graph. (c) At what value of t is the velocity zero? 8. (a) Use the data in Problem 1 to construct a smooth graph of position versus time. (b) By constructing tangents to the x(t) curve, find the instantaneous velocity of the car at several instants. (c) Plot the instantaneous velocity versus time and, from this, determine the average acceleration of the car. (d) What was the initial velocity of the car? 11. A 50.0-g superball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.50 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms ϭ 10Ϫ3 s.) 12. A particle starts from rest and accelerates as shown in Figure P2.12. Determine (a) the particle’s speed at t ϭ 10.0 s and at t ϭ 20.0 s, and (b) the distance traveled in the first 20.0 s. ax(m/s2) 2 1 t(s) 0 5 10 15 –1 –2 –3 Figure P2.12 20
    • Problems 13. Secretariat won the Kentucky Derby with times for successive quarter-mile segments of 25.2 s, 24.0 s, 23.8 s, and 23.0 s. (a) Find his average speed during each quarter-mile segment. (b) Assuming that Secretariat’s instantaneous speed at the finish line was the same as the average speed during the final quarter mile, find his average acceleration for the entire race. (Horses in the Derby start from rest.) 14. A velocity–time graph for an object moving along the x axis is shown in Figure P2.14. (a) Plot a graph of the acceleration versus time. (b) Determine the average acceleration of the object in the time intervals t ϭ 5.00 s to t ϭ 15.0 s and t ϭ 0 to t ϭ 20.0 s. 8 6 4 2 5 10 15 20 t(s) 18. Draw motion diagrams for (a) an object moving to the right at constant speed, (b) an object moving to the right and speeding up at a constant rate, (c) an object moving to the right and slowing down at a constant rate, (d) an object moving to the left and speeding up at a constant rate, and (e) an object moving to the left and slowing down at a constant rate. (f) How would your drawings change if the changes in speed were not uniform; that is, if the speed were not changing at a constant rate? Section 2.5 One-Dimensional Motion with Constant Acceleration 20. A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration. 21. Figure P2.14 15. Section 2.4 Motion Diagrams 19. Jules Verne in 1865 suggested sending people to the Moon by firing a space capsule from a 220-m-long cannon with a launch speed of 10.97 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during launch? Compare your answer with the free-fall acceleration 9.80 m/s2. vx(m/s) –2 –4 –6 –8 A particle moves along the x axis according to the equation x ϭ 2.00 ϩ 3.00t Ϫ 1.00t 2, where x is in meters and t is in seconds. At t ϭ 3.00 s, find (a) the position of the particle, (b) its velocity, and (c) its acceleration. 16. An object moves along the x axis according to the equation x(t) ϭ (3.00t 2 Ϫ 2.00t ϩ 3.00) m. Determine (a) the average speed between t ϭ 2.00 s and t ϭ 3.00 s, (b) the instantaneous speed at t ϭ 2.00 s and at t ϭ 3.00 s, (c) the average acceleration between t ϭ 2.00 s and t ϭ 3.00 s, and (d) the instantaneous acceleration at t ϭ 2.00 s and t ϭ 3.00 s. 17. Figure P2.17 shows a graph of vx versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. (a) Find the average acceleration for the time interval t ϭ 0 to t ϭ 6.00 s. (b) Estimate the time at which the acceleration has its greatest positive value and the value of the acceleration at that instant. (c) When is the acceleration zero? (d) Estimate the maximum negative value of the acceleration and the time at which it occurs. An object moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is Ϫ 5.00 cm, what is its acceleration? 22. A 745i BMW car can brake to a stop in a distance of 121 ft. from a speed of 60.0 mi/h. To brake to a stop from a speed of 80.0 mi/h requires a stopping distance of 211 ft. What is the average braking acceleration for (a) 60 mi/h to rest, (b) 80 mi/h to rest, (c) 80 mi/h to 60 mi/h? Express the answers in mi/h/s and in m/s2. 23. A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of Ϫ 3.50 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy? 24. Figure P2.24 represents part of the performance data of a car owned by a proud physics student. (a) Calculate from the graph the total distance traveled. (b) What distance does the car travel between the times t ϭ 10 s and t ϭ 40 s? (c) Draw a graph of its acceleration versus time between t ϭ 0 and t ϭ 50 s. (d) Write an equation for x as a function of time for each phase of the motion, represented by (i) 0a, (ii) ab, (iii) bc. (e) What is the average velocity of the car between t ϭ 0 and t ϭ 50 s? vx(m/s) a 50 vx(m/s) b 40 10 8 30 6 20 4 10 2 0 2 51 4 6 8 Figure P2.17 10 12 t(s) 0 10 20 30 40 Figure P2.24 c 50 t(s)
    • 52 C H A P T E R 2 • Motion in One Dimension 25. A particle moves along the x axis. Its position is given by the equation x ϭ 2 ϩ 3t Ϫ 4t 2 with x in meters and t in seconds. Determine (a) its position when it changes direction and (b) its velocity when it returns to the position it had at t ϭ 0. 26. In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.5 m/s. The driver of the Thunderbird realizes he must make a pit stop, and he smoothly slows to a stop over a distance of 250 m. He spends 5.00 s in the pit and then accelerates out, reaching his previous speed of 71.5 m/s after a distance of 350 m. At this point, how far has the Thunderbird fallen behind the Mercedes Benz, which has continued at a constant speed? 27. A jet plane lands with a speed of 100 m/s and can accelerate at a maximum rate of Ϫ 5.00 m/s2 as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time interval needed before it can come to rest? (b) Can this plane land on a small tropical island airport where the runway is 0.800 km long? 28. A car is approaching a hill at 30.0 m/s when its engine suddenly fails just at the bottom of the hill. The car moves with a constant acceleration of Ϫ 2.00 m/s2 while coasting up the hill. (a) Write equations for the position along the slope and for the velocity as functions of time, taking x ϭ 0 at the bottom of the hill, where vi ϭ 30.0 m/s. (b) Determine the maximum distance the car rolls up the hill. 29. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of Ϫ 5.60 m/s2 for 4.20 s, making straight skid marks 62.4 m long ending at the tree. With what speed does the car then strike the tree? 30. Help! One of our equations is missing! We describe constantacceleration motion with the variables and parameters vxi , vxf , ax , t , and xf Ϫ x i . Of the equations in Table 2.2, the first does not involve xf Ϫ x i . The second does not contain ax ; the third omits vxf and the last leaves out t. So to complete the set there should be an equation not involving vxi . Derive it from the others. Use it to solve Problem 29 in one step. 33. An electron in a cathode ray tube (CRT) accelerates from 2.00 ϫ 104 m/s to 6.00 ϫ 106 m/s over 1.50 cm. (a) How long does the electron take to travel this 1.50 cm? (b) What is its acceleration? 34. In a 100-m linear accelerator, an electron is accelerated to 1.00% of the speed of light in 40.0 m before it coasts for 60.0 m to a target. (a) What is the electron’s acceleration during the first 40.0 m? (b) How long does the total flight take? 35. Within a complex machine such as a robotic assembly line, suppose that one particular part glides along a straight track. A control system measures the average velocity of the part during each successive interval of time ⌬t0 ϭ t0 Ϫ 0, compares it with the value vc it should be, and switches a servo motor on and off to give the part a correcting pulse of acceleration. The pulse consists of a constant acceleration am applied for time interval ⌬tm ϭ tm Ϫ 0 within the next control time interval ⌬t0. As shown in Fig. P2.35, the part may be modeled as having zero acceleration when the motor is off (between tm and t0). A computer in the control system chooses the size of the acceleration so that the final velocity of the part will have the correct value vc . Assume the part is initially at rest and is to have instantaneous velocity vc at time t0. (a) Find the required value of am in terms of vc and tm . (b) Show that the displacement ⌬x of the part during the time interval ⌬t0 is given by ⌬x ϭ vc (t0 Ϫ 0.5tm). For specified values of vc and t0, (c) what is the minimum displacement of the part? (d) What is the maximum displacement of the part? (e) Are both the minimum and maximum displacements physically attainable? a am 0 tm t0 Figure P2.35 Photri, Inc. Courtesy U.S. Air Force 31. For many years Colonel John P. Stapp, USAF, held the world’s land speed record. On March 19, 1954, he rode a rocket-propelled sled that moved down a track at a speed of 632 mi/h. He and the sled were safely brought to rest in 1.40 s (Fig. P2.31). Determine (a) the negative acceleration he experienced and (b) the distance he traveled during this negative acceleration. 32. A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s. (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described? Figure P2.31 (Left) Col. John Stapp on rocket sled. (Right) Col. Stapp’s face is contorted by the stress of rapid negative acceleration. t
    • Problems 36. A glider on an air track carries a flag of length ᐉ through a stationary photogate, which measures the time interval ⌬td during which the flag blocks a beam of infrared light passing across the photogate. The ratio vd ϭ ᐉ/⌬td is the average velocity of the glider over this part of its motion. Suppose the glider moves with constant acceleration. (a) Argue for or against the idea that vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in space. (b) Argue for or against the idea that vd is equal to the instantaneous velocity of the glider when it is halfway through the photogate in time. 37. A ball starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 15.0 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane? (b) How long does it take to roll down the first plane? (c) What is the acceleration along the second plane? (d) What is the ball’s speed 8.00 m along the second plane? 53 42. A ball is thrown directly downward, with an initial speed of 8.00 m/s, from a height of 30.0 m. After what time interval does the ball strike the ground? 43. A student throws a set of keys vertically upward to her sorority sister, who is in a window 4.00 m above. The keys are caught 1.50 s later by the sister’s outstretched hand. (a) With what initial velocity were the keys thrown? (b) What was the velocity of the keys just before they were caught? 44. Emily challenges her friend David to catch a dollar bill as follows. She holds the bill vertically, as in Figure P2.44, with the center of the bill between David’s index finger and thumb. David must catch the bill after Emily releases it without moving his hand downward. If his reaction time is 0.2 s, will he succeed? Explain your reasoning. 38. Speedy Sue, driving at 30.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 155 m ahead traveling at 5.00 m/s. Sue applies her brakes but can accelerate only at Ϫ 2.00 m/s 2 because the road is wet. Will there be a collision? If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue’s car and the van. 39. Solve Example 2.8, “Watch out for the Speed Limit!” by a graphical method. On the same graph plot position versus time for the car and the police officer. From the intersection of the two curves read the time at which the trooper overtakes the car. Freely Falling Objects Note: In all problems in this section, ignore the effects of air resistance. George Semple Section 2.6 Figure P2.44 40. A golf ball is released from rest from the top of a very tall building. Neglecting air resistance, calculate (a) the position and (b) the velocity of the ball after 1.00, 2.00, and 3.00 s. 41. Every morning at seven o’clock There’s twenty terriers drilling on the rock. The boss comes around and he says, “Keep still And bear down heavy on the cast-iron drill And drill, ye terriers, drill.” And drill, ye terriers, drill. It’s work all day for sugar in your tea Down beyond the railway. And drill, ye terriers, drill. The foreman’s name was John McAnn. By God, he was a blamed mean man. One day a premature blast went off And a mile in the air went big Jim Goff. And drill ... Then when next payday came around Jim Goff a dollar short was found. When he asked what for, came this reply: “You were docked for the time you were up in the sky.” And drill... —American folksong What was Goff’s hourly wage? State the assumptions you make in computing it. 45. In Mostar, Bosnia, the ultimate test of a young man’s courage once was to jump off a 400-year-old bridge (now destroyed) into the River Neretva, 23.0 m below the bridge. (a) How long did the jump last? (b) How fast was the diver traveling upon impact with the water? (c) If the speed of sound in air is 340 m/s, how long after the diver took off did a spectator on the bridge hear the splash? 46. A ball is dropped from rest from a height h above the ground. Another ball is thrown vertically upwards from the ground at the instant the first ball is released. Determine the speed of the second ball if the two balls are to meet at a height h/2 above the ground. 47. A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height. Find (a) its initial velocity and (b) the height it reaches. 48. It is possible to shoot an arrow at a speed as high as 100 m/s. (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air?
    • 54 49. C H A P T E R 2 • Motion in One Dimension A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 10.0 m/s, and the distance from the limb to the level of the saddle is 3.00 m. (a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move? (b) How long is he in the air? 50. A woman is reported to have fallen 144 ft from the 17th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18.0 in. She suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventilator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box. 51. The height of a helicopter above the ground is given by h ϭ 3.00t 3, where h is in meters and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? 52. A freely falling object requires 1.50 s to travel the last 30.0 m before it hits the ground. From what height above the ground did it fall? Section 2.7 Kinematic Equations Derived from Calculus 53. Automotive engineers refer to the time rate of change of acceleration as the “jerk.” If an object moves in one dimension such that its jerk J is constant, (a) determine expressions for its acceleration ax(t), velocity vx(t), and position x(t), given that its initial acceleration, velocity, and position are axi , vxi , and xi , respectively. (b) Show that ax2 ϭ axi2 ϩ 2J(vx Ϫ vxi). 54. A student drives a moped along a straight road as described by the velocity-versus-time graph in Figure P2.54. Sketch this graph in the middle of a sheet of graph paper. (a) Directly above your graph, sketch a graph of the position versus time, aligning the time coordinates of the two graphs. (b) Sketch a graph of the acceleration versus time directly below the vx-t graph, again aligning the time coordinates. On each graph, show the numerical values of x and ax for all points of inflection. (c) What is the acceleration at t ϭ 6 s? (d) Find the position (relative to the starting point) at t ϭ 6 s. (e) What is the moped’s final position at t ϭ 9 s? vx(m/s) 8 6 4 2 –2 1 2 3 4 5 6 7 –4 –6 –8 Figure P2.54 8 9 10 t(s) 55. The speed of a bullet as it travels down the barrel of a rifle toward the opening is given by v ϭ (Ϫ 5.00 ϫ 107)t 2 ϩ (3.00 ϫ 105)t, where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero. (a) Determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel. (b) Determine the length of time the bullet is accelerated. (c) Find the speed at which the bullet leaves the barrel. (d) What is the length of the barrel? 56. The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given (in SI units) by a ϭ Ϫ 3.00v 2 for v Ͼ 0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble’s speed is reduced to half of its initial value? Additional Problems 57. A car has an initial velocity v0 when the driver sees an obstacle in the road in front of him. His reaction time is ⌬tr , and the braking acceleration of the car is a. Show that the total stopping distance is sstop ϭ v0 ⌬t r Ϫ v 02/2a. Remember that a is a negative number. 58. The yellow caution light on a traffic signal should stay on long enough to allow a driver to either pass through the intersection or safely stop before reaching the intersection. A car can stop if its distance from the intersection is greater than the stopping distance found in the previous problem. If the car is less than this stopping distance from the intersection, the yellow light should stay on long enough to allow the car to pass entirely through the intersection. (a) Show that the yellow light should stay on for a time interval ⌬t light ϭ ⌬t r Ϫ (v0/2a) ϩ (si /v0) where ⌬tr is the driver’s reaction time, v0 is the velocity of the car approaching the light at the speed limit, a is the braking acceleration, and si is the width of the intersection. (b) As city traffic planner, you expect cars to approach an intersection 16.0 m wide with a speed of 60.0 km/h. Be cautious and assume a reaction time of 1.10 s to allow for a driver’s indecision. Find the length of time the yellow light should remain on. Use a braking acceleration of Ϫ 2.00 m/s2. 59. The Acela is the Porsche of American trains. Shown in Figure P2.59a, the electric train whose name is pronounced ah-SELL-ah is in service on the Washington-New YorkBoston run. With two power cars and six coaches, it can carry 304 passengers at 170 mi/h. The carriages tilt as much as 6Њ from the vertical to prevent passengers from feeling pushed to the side as they go around curves. Its braking mechanism uses electric generators to recover its energy of motion. A velocity-time graph for the Acela is shown in Figure P2.59b. (a) Describe the motion of the train in each successive time interval. (b) Find the peak positive acceleration of the train in the motion graphed. (c) Find the train’s displacement in miles between t ϭ 0 and t ϭ 200 s.
    • Problems Courtesy Amtrak Nec Media Relations 65. Setting a new world record in a 100-m race, Maggie and Judy cross the finish line in a dead heat, both taking 10.2 s. Accelerating uniformly, Maggie took 2.00 s and Judy 3.00 s to attain maximum speed, which they maintained for the rest of the race. (a) What was the acceleration of each sprinter? (b) What were their respective maximum speeds? (c) Which sprinter was ahead at the 6.00-s mark, and by how much? (a) 200 v(mi/h) 150 100 50 0 –50 –50 55 0 50 100 150 200 250 300 350 400 t(s) –100 (b) Figure P2.59 (a) The Acela—1 171 000 lb of cold steel thundering along at 150 mi/h. (b) Velocity-versus-time graph for the Acela. 60. Liz rushes down onto a subway platform to find her train already departing. She stops and watches the cars go by. Each car is 8.60 m long. The first moves past her in 1.50 s and the second in 1.10 s. Find the constant acceleration of the train. 61. A dog’s hair has been cut and is now getting 1.04 mm longer each day. With winter coming on, this rate of hair growth is steadily increasing, by 0.132 mm/day every week. By how much will the dog’s hair grow during 5 weeks? 62. A test rocket is fired vertically upward from a well. A catapult gives it an initial speed of 80.0 m/s at ground level. Its engines then fire and it accelerates upward at 4.00 m/s2 until it reaches an altitude of 1 000 m. At that point its engines fail and the rocket goes into free fall, with an acceleration of Ϫ 9.80 m/s2. (a) How long is the rocket in motion above the ground? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the Earth? (You will need to consider the motion while the engine is operating separate from the free-fall motion.) 66. A commuter train travels between two downtown stations. Because the stations are only 1.00 km apart, the train never reaches its maximum possible cruising speed. During rush hour the engineer minimizes the time interval ⌬t between two stations by accelerating for a time interval ⌬t1 at a rate a1 ϭ 0.100 m/s2 and then immediately braking with acceleration a 2 ϭ Ϫ 0.500 m/s2 for a time interval ⌬t 2. Find the minimum time interval of travel ⌬t and the time interval ⌬t1. 67. A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Suppose that the maximum depth of the dent is on the order of 1 cm. Compute an order-of-magnitude estimate for the maximum acceleration of the ball while it is in contact with the pavement. State your assumptions, the quantities you estimate, and the values you estimate for them. 68. At NASA’s John H. Glenn research center in Cleveland, Ohio, free-fall research is performed by dropping experiment packages from the top of an evacuated shaft 145 m high. Free fall imitates the so-called microgravity environment of a satellite in orbit. (a) What is the maximum time interval for free fall if an experiment package were to fall the entire 145 m? (b) Actual NASA specifications allow for a 5.18-s drop time interval. How far do the packages drop and (c) what is their speed at 5.18 s? (d) What constant acceleration would be required to stop an experiment package in the distance remaining in the shaft after its 5.18-s fall? 69. An inquisitive physics student and mountain climber climbs a 50.0-m cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.00 m/s. (a) How long after release of the first stone do the two stones hit the water? (b) What initial velocity must the second stone have if they are to hit simultaneously? (c) What is the speed of each stone at the instant the two hit the water? 63. A motorist drives along a straight road at a constant speed of 15.0 m/s. Just as she passes a parked motorcycle police officer, the officer starts to accelerate at 2.00 m/s2 to overtake her. Assuming the officer maintains this acceleration, (a) determine the time it takes the police officer to reach the motorist. Find (b) the speed and (c) the total displacement of the officer as he overtakes the motorist. 70. A rock is dropped from rest into a well. The well is not really 16 seconds deep, as in Figure P2.70. (a) The sound of the splash is actually heard 2.40 s after the rock is released from rest. How far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is 336 m/s. (b) What If? If the travel time for the sound is neglected, what percentage error is introduced when the depth of the well is calculated? 64. In Figure 2.10b, the area under the velocity versus time curve and between the vertical axis and time t (vertical dashed line) represents the displacement. As shown, this area consists of a rectangle and a triangle. Compute their areas and compare the sum of the two areas with the expression on the right-hand side of Equation 2.12. 71. To protect his food from hungry bears, a boy scout raises his food pack with a rope that is thrown over a tree limb at height h above his hands. He walks away from the vertical rope with constant velocity v boy, holding the free end of the rope in his hands (Fig. P2.71). (a) Show that the speed v of the food pack is given by x(x 2 ϩ h 2)Ϫ1/2 v boy where x
    • 56 C H A P T E R 2 • Motion in One Dimension By permission of John Hart and Creators Syndicate, Inc. Figure P2.70 is the distance he has walked away from the vertical rope. (b) Show that the acceleration a of the food pack is h2(x 2 ϩ h2)Ϫ3/2 v 2 . (c) What values do the acceleration a boy and velocity v have shortly after he leaves the point under the pack (x ϭ 0)? (d) What values do the pack’s velocity and acceleration approach as the distance x continues to increase? 74. Astronauts on a distant planet toss a rock into the air. With the aid of a camera that takes pictures at a steady rate, they record the height of the rock as a function of time as given in Table P2.74. (a) Find the average velocity of the rock in the time interval between each measurement and the next. (b) Using these average velocities to approximate instantaneous velocities at the midpoints of the time intervals, make a graph of velocity as a function of time. Does the rock move with constant acceleration? If so, plot a straight line of best fit on the graph and calculate its slope to find the acceleration. Table P2.74 Height of a Rock versus Time v a Time (s) h m v boy x Height (m) Time (s) Height (m) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 5.00 5.75 6.40 6.94 7.38 7.72 7.96 8.10 8.13 8.07 7.90 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 7.62 7.25 6.77 6.20 5.52 4.73 3.85 2.86 1.77 0.58 Figure P2.71 Problems 71 and 72. 72. In Problem 71, let the height h equal 6.00 m and the speed v boy equal 2.00 m/s. Assume that the food pack starts from rest. (a) Tabulate and graph the speed–time graph. (b) Tabulate and graph the acceleration-time graph. Let the range of time be from 0 s to 5.00 s and the time intervals be 0.500 s. 73. Kathy Kool buys a sports car that can accelerate at the rate of 4.90 m/s2. She decides to test the car by racing with another speedster, Stan Speedy. Both start from rest, but experienced Stan leaves the starting line 1.00 s before Kathy. If Stan moves with a constant acceleration of 3.50 m/s2 and Kathy maintains an acceleration of 4.90 m/s2, find (a) the time at which Kathy overtakes Stan, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant she overtakes him. 75. Two objects, A and B, are connected by a rigid rod that has a length L. The objects slide along perpendicular guide rails, as shown in Figure P2.75. If A slides to the left with a constant speed v, find the velocity of B when ␣ ϭ 60.0°. y B x L y α O v A x Figure P2.75
    • Answers to Quick Quizzes Answers to Quick Quizzes 2.1 (c). If the particle moves along a line without changing direction, the displacement and distance traveled over any time interval will be the same. As a result, the magnitude of the average velocity and the average speed will be the same. If the particle reverses direction, however, the displacement will be less than the distance traveled. In turn, the magnitude of the average velocity will be smaller than the average speed. 57 2.4 (c). If a particle with constant acceleration stops and its acceleration remains constant, it must begin to move again in the opposite direction. If it did not, the acceleration would change from its original constant value to zero. Choice (a) is not correct because the direction of acceleration is not specified by the direction of the velocity. Choice (b) is also not correct by counterexample—a car moving in the Ϫ x direction and slowing down has a positive acceleration. 4.0 2.5 Graph (a) has a constant slope, indicating a constant acceleration; this is represented by graph (e). Graph (b) represents a speed that is increasing constantly but not at a uniform rate. Thus, the acceleration must be increasing, and the graph that best indicates this is (d). Graph (c) depicts a velocity that first increases at a constant rate, indicating constant acceleration. Then the velocity stops increasing and becomes constant, indicating zero acceleration. The best match to this situation is graph (f). 2.0 2.6 (e). For the entire time interval that the ball is in free fall, the acceleration is that due to gravity. 2.2 (b). If the car is slowing down, a force must be pulling in the direction opposite to its velocity. 2.3 False. Your graph should look something like the following. This vx -t graph shows that the maximum speed is about 5.0 m/s, which is 18 km/h (ϭ 11 mi/h), and so the driver was not speeding. 6.0 vx(m/s) 0.0 10 20 30 –2.0 40 t(s) 50 2.7 (d). While the ball is rising, it is slowing down. After reaching the highest point, the ball begins to fall and its speed increases. 2.8 (a). At the highest point, the ball is momentarily at rest, but still accelerating at Ϫ g. –4.0 –6.0 Answer to Quick Quiz 2.3
    • Chapter 3 Vectors CHAPTE R OUTLI N E 3.1 Coordinate Systems 3.2 Vector and Scalar Quantities 3.3 Some Properties of Vectors 3.4 Components of a Vector and Unit Vectors L These controls in the cockpit of a commercial aircraft assist the pilot in maintaining 58 control over the velocity of the aircraft—how fast it is traveling and in what direction it is traveling—allowing it to land safely. Quantities that are defined by both a magnitude and a direction, such as velocity, are called vector quantities. (Mark Wagner/Getty Images)
    • In our study of physics, we often need to work with physical quantities that have both numerical and directional properties. As noted in Section 2.1, quantities of this nature are vector quantities. This chapter is primarily concerned with vector algebra and with some general properties of vector quantities. We discuss the addition and subtraction of vector quantities, together with some common applications to physical situations. Vector quantities are used throughout this text, and it is therefore imperative that you master both their graphical and their algebraic properties. 3.1 Coordinate Systems y Many aspects of physics involve a description of a location in space. In Chapter 2, for example, we saw that the mathematical description of an object’s motion requires a method for describing the object’s position at various times. This description is accomplished with the use of coordinates, and in Chapter 2 we used the Cartesian coordinate system, in which horizontal and vertical axes intersect at a point defined as the origin (Fig. 3.1). Cartesian coordinates are also called rectangular coordinates. Sometimes it is more convenient to represent a point in a plane by its plane polar coordinates (r, ␪), as shown in Figure 3.2a. In this polar coordinate system, r is the distance from the origin to the point having Cartesian coordinates (x, y), and ␪ is the angle between a line drawn from the origin to the point and a fixed axis. This fixed axis is usually the positive x axis, and ␪ is usually measured counterclockwise from it. From the right triangle in Figure 3.2b, we find that sin ␪ ϭ y/r and that cos ␪ ϭ x/r. (A review of trigonometric functions is given in Appendix B.4.) Therefore, starting with the plane polar coordinates of any point, we can obtain the Cartesian coordinates by using the equations x ϭ r cos ␪ (3.1) y ϭ r sin ␪ (x, y) r θ x O (a) y sin θ = r cos θ = x θ r tan θ = r y y x (3.2) θ x (b) Active Figure 3.2 (a) The plane polar coordinates of a point are represented by the distance r and the angle ␪, where ␪ is measured counterclockwise from the positive x axis. (b) The right triangle used to relate (x, y) to (r, ␪). y (x, y) Q P (–3, 4) (5, 3) O x Figure 3.1 Designation of points in a Cartesian coordinate system. Every point is labeled with coordinates (x, y). At the Active Figures link at http://www.pse6.com, you can move the point and see the changes to the rectangular and polar coordinates as well as to the sine, cosine, and tangent of angle ␪. 59
    • 60 C HAPTE R 3 • Vectors Furthermore, the definitions of trigonometry tell us that tan ␪ ϭ y x (3.3) r ϭ √x 2 ϩ y 2 (3.4) Equation 3.4 is the familiar Pythagorean theorem. These four expressions relating the coordinates (x, y) to the coordinates (r, ␪) apply only when ␪ is defined as shown in Figure 3.2a— in other words, when positive ␪ is an angle measured counterclockwise from the positive x axis. (Some scientific calculators perform conversions between Cartesian and polar coordinates based on these standard conventions.) If the reference axis for the polar angle ␪ is chosen to be one other than the positive x axis or if the sense of increasing ␪ is chosen differently, then the expressions relating the two sets of coordinates will change. Example 3.1 Polar Coordinates The Cartesian coordinates of a point in the xy plane are (x, y) ϭ (Ϫ 3.50, Ϫ 2.50) m, as shown in Figure 3.3. Find the polar coordinates of this point. Solution For the examples in this and the next two chapters we will illustrate the use of the General Problem-Solving y(m) Strategy outlined at the end of Chapter 2. In subsequent chapters, we will make fewer explicit references to this strategy, as you will have become familiar with it and should be applying it on your own. The drawing in Figure 3.3 helps us to conceptualize the problem. We can categorize this as a plugin problem. From Equation 3.4, r ϭ √x 2 ϩ y 2 ϭ √(Ϫ3.50 m)2 ϩ (Ϫ2.50 m)2 ϭ 4.30 m and from Equation 3.3, θ x(m) tan ␪ ϭ r (–3.50, –2.50) y Ϫ2.50 m ϭ ϭ 0.714 x Ϫ3.50 m ␪ ϭ 216Њ Active Figure 3.3 (Example 3.1) Finding polar coordinates when Cartesian coordinates are given. At the Active Figures link at http://www.pse6.com, you can move the point in the xy plane and see how its Cartesian and polar coordinates change. Note that you must use the signs of x and y to find that the point lies in the third quadrant of the coordinate system. That is, ␪ ϭ 216° and not 35.5°. 3.2 Vector and Scalar Quantities As noted in Chapter 2, some physical quantities are scalar quantities whereas others are vector quantities. When you want to know the temperature outside so that you will know how to dress, the only information you need is a number and the unit “degrees C” or “degrees F.” Temperature is therefore an example of a scalar quantity: A scalar quantity is completely specified by a single value with an appropriate unit and has no direction. Other examples of scalar quantities are volume, mass, speed, and time intervals. The rules of ordinary arithmetic are used to manipulate scalar quantities. If you are preparing to pilot a small plane and need to know the wind velocity, you must know both the speed of the wind and its direction. Because direction is important for its complete specification, velocity is a vector quantity:
    • S ECTI O N 3.3 • Some Properties of Vectors Ꭾ A vector quantity is completely specified by a number and appropriate units plus a direction. Another example of a vector quantity is displacement, as you know from Chapter 2. Suppose a particle moves from some point Ꭽ to some point Ꭾ along a straight path, as shown in Figure 3.4. We represent this displacement by drawing an arrow from Ꭽ to Ꭾ, with the tip of the arrow pointing away from the starting point. The direction of the arrowhead represents the direction of the displacement, and the length of the arrow represents the magnitude of the displacement. If the particle travels along some other path from Ꭽ to Ꭾ, such as the broken line in Figure 3.4, its displacement is still the arrow drawn from Ꭽ to Ꭾ. Displacement depends only on the initial and final positions, so the displacement vector is independent of the path taken between these two points. In this text, we use a boldface letter, such as A, to represent a vector quantity. Another notation is useful when boldface notation is difficult, such as when writing on pa: per or on a chalkboard—an arrow is written over the symbol for the vector: A. The magnitude of the vector A is written either A or ͉A͉. The magnitude of a vector has physical units, such as meters for displacement or meters per second for velocity. The magnitude of a vector is always a positive number. Ꭽ Figure 3.4 As a particle moves from Ꭽ to Ꭾ along an arbitrary path represented by the broken line, its displacement is a vector quantity shown by the arrow drawn from Ꭽ to Ꭾ. y Quick Quiz 3.1 O Which of the following are vector quantities and which are scalar quantities? (a) your age (b) acceleration (c) velocity (d) speed (e) mass 3.3 Some Properties of Vectors x Figure 3.5 These four vectors are equal because they have equal lengths and point in the same direction. Equality of Two Vectors For many purposes, two vectors A and B may be defined to be equal if they have the same magnitude and point in the same direction. That is, A ϭ B only if A ϭ B and if A and B point in the same direction along parallel lines. For example, all the vectors in Figure 3.5 are equal even though they have different starting points. This property allows us to move a vector to a position parallel to itself in a diagram without affecting the vector. Adding Vectors The rules for adding vectors are conveniently described by graphical methods. To add vector B to vector A, first draw vector A on graph paper, with its magnitude represented by a convenient length scale, and then draw vector B to the same scale with its tail starting from the tip of A, as shown in Figure 3.6. The resultant vector R ϭ A ϩ B is the vector drawn from the tail of A to the tip of B. For example, if you walked 3.0 m toward the east and then 4.0 m toward the north, as shown in Figure 3.7, you would find yourself 5.0 m from where you started, measured at an angle of 53° north of east. Your total displacement is the vector sum of the individual displacements. A geometric construction can also be used to add more than two vectors. This is shown in Figure 3.8 for the case of four vectors. The resultant vector R ϭ A ϩ B ϩ C ϩ D is the vector that completes the polygon. In other words, R is the vector drawn from the tail of the first vector to the tip of the last vector. When two vectors are added, the sum is independent of the order of the addition. (This fact may seem trivial, but as you will see in Chapter 11, the order is important 61 R = A + B B A Active Figure 3.6 When vector B is added to vector A, the resultant R is the vector that runs from the tail of A to the tip of B. Go to the Active Figures link at http://www.pse6.com. L PITFALL PREVENTION 3.1 Vector Addition versus Scalar Addition Keep in mind that A ϩ B ϭ C is very different from A ϩ B ϭ C. The first is a vector sum, which must be handled carefully, such as with the graphical method described here. The second is a simple algebraic addition of numbers that is handled with the normal rules of arithmetic.
    • C HAPTE R 3 • Vectors 62 N E m W +( 4.0 m )2 =5 .0 S |R |= (3 .0 m )2 4.0 m ( 4.0 ) = 53° 3.0 θ = tan–1 θ Figure 3.7 Vector addition. Walking first 3.0 m due east and then 4.0 m due north leaves you 5.0 m from your starting point. 3.0 m when vectors are multiplied). This can be seen from the geometric construction in Figure 3.9 and is known as the commutative law of addition: AϩBϭBϩA + C + D D When three or more vectors are added, their sum is independent of the way in which the individual vectors are grouped together. A geometric proof of this rule for three vectors is given in Figure 3.10. This is called the associative law of addition: = A + B C (3.5) R A ϩ (B ϩ C) ϭ (A ϩ B) ϩ C B A Figure 3.8 Geometric construction for summing four vectors. The resultant vector R is by definition the one that completes the polygon. (3.6) In summary, a vector quantity has both magnitude and direction and also obeys the laws of vector addition as described in Figures 3.6 to 3.10. When two or more vectors are added together, all of them must have the same units and all of them must be the same type of quantity. It would be meaningless to add a velocity vector (for example, 60 km/h to the east) to a displacement vector (for example, 200 km to the north) because they represent different physical quantities. The same rule also applies to scalars. For example, it would be meaningless to add time intervals to temperatures. Negative of a Vector The negative of the vector A is defined as the vector that when added to A gives zero for the vector sum. That is, A ϩ (Ϫ A) ϭ 0. The vectors A and Ϫ A have the same magnitude but point in opposite directions. B =B Associative Law R B +A =A +B A C B) + (B Figure 3.9 This construction shows that A ϩ B ϭ B ϩ A—in other words, that vector addition is commutative. + (A + B+C A A C + C) C A+B B A B A Figure 3.10 Geometric constructions for verifying the associative law of addition.
    • S ECTI O N 3.3 • Some Properties of Vectors Vector Subtraction B A –B C=A–B C=A–B B A (b) (a) Figure 3.11 (a) This construction shows how to subtract vector B from vector A. The vector Ϫ B is equal in magnitude to vector B and points in the opposite direction. To subtract B from A, apply the rule of vector addition to the combination of A and Ϫ B: Draw A along some convenient axis, place the tail of Ϫ B at the tip of A, and C is the difference A Ϫ B. (b) A second way of looking at vector subtraction. The difference vector C ϭ A Ϫ B is the vector that we must add to B to obtain A. Subtracting Vectors The operation of vector subtraction makes use of the definition of the negative of a vector. We define the operation A Ϫ B as vector Ϫ B added to vector A: A Ϫ B ϭ A ϩ (Ϫ B) (3.7) The geometric construction for subtracting two vectors in this way is illustrated in Figure 3.11a. Another way of looking at vector subtraction is to note that the difference A Ϫ B between two vectors A and B is what you have to add to the second vector to obtain the first. In this case, the vector A Ϫ B points from the tip of the second vector to the tip of the first, as Figure 3.11b shows. Quick Quiz 3.2 The magnitudes of two vectors A and B are A ϭ 12 units and B ϭ 8 units. Which of the following pairs of numbers represents the largest and smallest possible values for the magnitude of the resultant vector R ϭ A ϩ B? (a) 14.4 units, 4 units (b) 12 units, 8 units (c) 20 units, 4 units (d) none of these answers. Quick Quiz 3.3 If vector B is added to vector A, under what condition does the resultant vector A ϩ B have magnitude A ϩ B? (a) A and B are parallel and in the same direction. (b) A and B are parallel and in opposite directions. (c) A and B are perpendicular. Quick Quiz 3.4 If vector B is added to vector A, which two of the following choices must be true in order for the resultant vector to be equal to zero? (a) A and B are parallel and in the same direction. (b) A and B are parallel and in opposite directions. (c) A and B have the same magnitude. (d) A and B are perpendicular. 63
    • 64 C HAPTE R 3 • Vectors Example 3.2 A Vacation Trip A car travels 20.0 km due north and then 35.0 km in a direction 60.0° west of north, as shown in Figure 3.12a. Find the magnitude and direction of the car’s resultant displacement. Solution The vectors A and B drawn in Figure 3.12a help us to conceptualize the problem. We can categorize this as a relatively simple analysis problem in vector addition. The displacement R is the resultant when the two individual displacements A and B are added. We can further categorize this as a problem about the analysis of triangles, so we appeal to our expertise in geometry and trigonometry. In this example, we show two ways to analyze the problem of finding the resultant of two vectors. The first way is to solve the problem geometrically, using graph paper and a protractor to measure the magnitude of R and its direction in Figure 3.12a. (In fact, even when you know you are going to be carrying out a calculation, you should sketch the vectors to check your results.) With an ordinary ruler and protractor, a large diagram typically gives answers to two-digit but not to three-digit precision. The second way to solve the problem is to analyze it algebraically. The magnitude of R can be obtained from the law of cosines as applied to the triangle (see Appendix B.4). With ␪ ϭ 180° Ϫ 60° ϭ 120° and R 2 ϭ A2 ϩ B 2 Ϫ 2AB cos ␪, we find that R ϭ √A2 ϩ B 2 Ϫ 2AB cos ␪ ϭ √(20.0 km)2 ϩ (35.0 km)2 Ϫ 2(20.0 km)(35.0 km) cos 120Њ ϭ 48.2 km The direction of R measured from the northerly direction can be obtained from the law of sines (Appendix B.4): sin ␤ sin ␪ ϭ B R sin ␤ ϭ 〉 35.0 km sin ␪ ϭ sin 120Њ ϭ 0.629 R 48.2 km ␤ ϭ 39.0Њ The resultant displacement of the car is 48.2 km in a direction 39.0° west of north. We now finalize the problem. Does the angle ␤ that we calculated agree with an estimate made by looking at Figure 3.12a or with an actual angle measured from the diagram using the graphical method? Is it reasonable that the magnitude of R is larger than that of both A and B? Are the units of R correct? While the graphical method of adding vectors works well, it suffers from two disadvantages. First, some individuals find using the laws of cosines and sines to be awkward. Second, a triangle only results if you are adding two vectors. If you are adding three or more vectors, the resulting geometric shape is not a triangle. In Section 3.4, we explore a new method of adding vectors that will address both of these disadvantages. What If? Suppose the trip were taken with the two vectors in reverse order: 35.0 km at 60.0° west of north first, and then 20.0 km due north. How would the magnitude and the direction of the resultant vector change? Answer They would not change. The commutative law for vector addition tells us that the order of vectors in an addition is irrelevant. Graphically, Figure 3.12b shows that the vectors added in the reverse order give us the same resultant vector. N W E S y(km) y(km) 40 B R 60.0° A 20 θ β –20 40 R 20 B A 0 (a) x(km) β –20 0 (b) x(km) Figure 3.12 (Example 3.2) (a) Graphical method for finding the resultant displacement vector R ϭ A ϩ B. (b) Adding the vectors in reverse order (B ϩ A) gives the same result for R.
    • S ECTI O N 3.4 • Components of a Vector and Unit Vectors y Multiplying a Vector by a Scalar If vector A is multiplied by a positive scalar quantity m, then the product mA is a vector that has the same direction as A and magnitude mA. If vector A is multiplied by a negative scalar quantity Ϫ m, then the product Ϫ mA is directed opposite A. For example, the vector 5A is five times as long as A and points in the same direction as A; the vector Ϫ 1A is one-third the length of A and points in the direction opposite A. 3 3.4 θ O y Ax ϭ A cos ␪ (3.8) Ay ϭ A sin ␪ (3.9) These components form two sides of a right triangle with a hypotenuse of length A. Thus, it follows that the magnitude and direction of A are related to its components through the expressions A ϭ √Ax2 ϩ Ay2 (3.10) A ΂A ΃ (3.11) x x Ax (a) The graphical method of adding vectors is not recommended whenever high accuracy is required or in three-dimensional problems. In this section, we describe a method of adding vectors that makes use of the projections of vectors along coordinate axes. These projections are called the components of the vector. Any vector can be completely described by its components. Consider a vector A lying in the xy plane and making an arbitrary angle ␪ with the positive x axis, as shown in Figure 3.13a. This vector can be expressed as the sum of two other vectors Ax and Ay . From Figure 3.13b, we see that the three vectors form a right triangle and that A ϭ Ax ϩ Ay . We shall often refer to the “components of a vector A,” written Ax and Ay (without the boldface notation). The component Ax represents the projection of A along the x axis, and the component Ay represents the projection of A along the y axis. These components can be positive or negative. The component Ax is positive if Ax points in the positive x direction and is negative if Ax points in the negative x direction. The same is true for the component Ay . From Figure 3.13 and the definition of sine and cosine, we see that cos ␪ ϭ Ax /A and that sin ␪ ϭ Ay /A. Hence, the components of A are y A Ay Components of a Vector and Unit Vectors ␪ ϭ tan Ϫ1 65 Note that the signs of the components Ax and Ay depend on the angle ␪. For example, if ␪ ϭ 120°, then Ax is negative and Ay is positive. If ␪ ϭ 225°, then both Ax and Ay are negative. Figure 3.14 summarizes the signs of the components when A lies in the various quadrants. When solving problems, you can specify a vector A either with its components Ax and Ay or with its magnitude and direction A and ␪. Quick Quiz 3.5 Choose the correct response to make the sentence true: A component of a vector is (a) always, (b) never, or (c) sometimes larger than the magnitude of the vector. A Ay θ O x Ax (b) Figure 3.13 (a) A vector A lying in the xy plane can be represented by its component vectors Ax and Ay . (b) The y component vector Ay can be moved to the right so that it adds to Ax . The vector sum of the component vectors is A. These three vectors form a right triangle. Components of the vector A L PITFALL PREVENTION 3.2 Component Vectors versus Components The vectors Ax and Ay are the component vectors of A. These should not be confused with the scalars Ax and Ay , which we shall always refer to as the components of A. y Ax positive Ay positive Ay positive Ax negative Suppose you are working a physics problem that requires resolving a vector into its components. In many applications it is convenient to express the components in a coordinate system having axes that are not horizontal and vertical but are still perpendicular to each other. If you choose reference axes or an angle other than the axes and angle shown in Figure 3.13, the components must be modified accordingly. Suppose a Ax negative Ax positive Ay negative Ay negative x Figure 3.14 The signs of the components of a vector A depend on the quadrant in which the vector is located.
    • C HAPTE R 3 • Vectors 66 y′ x′ B By′ θ′ Bx′ Unit Vectors O Figure 3.15 The component vectors of B in a coordinate system that is tilted. y x ˆ j ˆ i ˆ k y For example, consider a point lying in the xy plane and having Cartesian coordinates (x, y), as in Figure 3.17. The point can be specified by the position vector r, which in unit–vector form is given by (3.13) r ϭ xˆ ϩ yˆ i j (a) y A Axˆ i Vector quantities often are expressed in terms of unit vectors. A unit vector is a dimensionless vector having a magnitude of exactly 1. Unit vectors are used to specify a given direction and have no other physical significance. They are used solely as a conveˆ nience in describing a direction in space. We shall use the symbols ˆ , ˆ and k to reprei j, sent unit vectors pointing in the positive x, y, and z directions, respectively. (The “hats” on ˆ the symbols are a standard notation for unit vectors.) The unit vectors ˆ , ˆ and k form a i j, set of mutually perpendicular vectors in a right-handed coordinate system, as shown in Figure 3.16a. The magnitude of each unit vector equals 1; that is, ͉ ˆ = ͉ ˆ = ͉ k͉ = 1. i͉ j͉ ˆ Consider a vector A lying in the xy plane, as shown in Figure 3.16b. The product i i of the component Ax and the unit vector ˆ is the vector Axˆ , which lies on the x axis i and has magnitude ͉Ax ͉. (The vector Ax ˆ is an alternative representation of vector Ax.) Likewise, Ayˆ is a vector of magnitude ͉Ay ͉ lying on the y axis. (Again, vector Ayˆ j j is an alternative representation of vector Ay.) Thus, the unit–vector notation for the vector A is i j (3.12) AϭA ˆ ϩA ˆ x z Ay ˆ j vector B makes an angle ␪Ј with the xЈ axis defined in Figure 3.15. The components of B along the xЈ and yЈ axes are BxЈ ϭ B cos ␪Ј and ByЈ ϭ B sin ␪Ј, as specified by Equations 3.8 and 3.9. The magnitude and direction of B are obtained from expressions equivalent to Equations 3.10 and 3.11. Thus, we can express the components of a vector in any coordinate system that is convenient for a particular situation. x (b) Active Figure 3.16 (a) The unit ˆ vectors ˆ , ˆ and k are directed i j, along the x, y, and z axes, respeci j tively. (b) Vector A ϭ Ax ˆ ϩ Ay ˆ lying in the xy plane has components Ax and Ay. This notation tells us that the components of r are the lengths x and y. Now let us see how to use components to add vectors when the graphical method is not sufficiently accurate. Suppose we wish to add vector B to vector A in Equation 3.12, where vector B has components Bx and By. All we do is add the x and y components separately. The resultant vector R ϭ A ϩ B is therefore i j) i j) R ϭ (A ˆ ϩ A ˆ ϩ (B ˆ ϩ B ˆ x y x y or R ϭ (Ax ϩ Bx)ˆ ϩ (Ay ϩ By)ˆ i j (3.14) Because R ϭ Rx ˆ ϩ Ry ˆ we see that the components of the resultant vector are i j, Rx ϭ Ax ϩ Bx (3.15) Ry ϭ Ay ϩ By At the Active Figures link at http://www.pse6.com you can rotate the coordinate axes in 3-dimensional space and view a representation of vector A in three dimensions. y (x,y) r Figure 3.17 The point whose Cartesian coordinates are (x, y) can be represented by the position vector r ϭ xˆ ϩ yˆ i j. O x
    • S ECTI O N 3.4 • Components of a Vector and Unit Vectors y We obtain the magnitude of R and the angle it makes with the x axis from its components, using the relationships R ϭ √R x2 ϩ R y2 ϭ √(Ax ϩ Bx)2 ϩ (Ay ϩ By)2 Ry Ay ϩ By tan ␪ ϭ ϭ Rx Ax ϩ Bx (3.16) (3.17) We can check this addition by components with a geometric construction, as shown in Figure 3.18. Remember that you must note the signs of the components when using either the algebraic or the graphical method. At times, we need to consider situations involving motion in three component directions. The extension of our methods to three-dimensional vectors is straightforward. If A and B both have x, y, and z components, we express them in the form ˆ i j (3.18) AϭA ˆ ϩA ˆϩA k x y z ˆ B ϭ Bx ˆ ϩ By ˆ ϩ Bz k i j (3.19) R ϭ (Ax ϩ Bx )ˆ ϩ (Ay ϩ By )ˆ ϩ (Az ϩ Bz)ˆ i j k 67 Ry R By Ay B A Ax x Bx Rx Figure 3.18 This geometric construction for the sum of two vectors shows the relationship between the components of the resultant R and the components of the individual vectors. (3.20) The sum of A and B is Note that Equation 3.20 differs from Equation 3.14: in Equation 3.20, the resultant vector also has a z component R z ϭ Az ϩ B z . If a vector R has x, y, and z components, the magnitude of the vector is R ϭ √R x2 ϩ R y 2 ϩ R z 2. The angle ␪x that R makes with the x axis is found from the expression cos ␪x ϭ Rx /R, with similar expressions for the angles with respect to the y and z axes. L Quick Quiz 3.6 If at least one component of a vector is a positive number, the vector cannot (a) have any component that is negative (b) be zero (c) have three dimensions. Quick Quiz 3.7 If A ϩ B ϭ 0, the corresponding components of the two vectors A and B must be (a) equal (b) positive (c) negative (d) of opposite sign. Quick Quiz 3.8 For which of the following vectors is the magnitude of the ˆ vector equal to one of the components of the vector? (a) A ϭ 2i ϩ 5ˆ (b) B ϭ Ϫ 3ˆ j j ˆ (c) C ϭ ϩ 5 k P R O B L E M - S O LV I N G H I N T S Adding Vectors When you need to add two or more vectors, use this step-by-step procedure: • • • • Select a coordinate system that is convenient. (Try to reduce the number of components you need to calculate by choosing axes that line up with as many vectors as possible.) Draw a labeled sketch of the vectors described in the problem. Find the x and y components of all vectors and the resultant components (the algebraic sum of the components) in the x and y directions. If necessary, use the Pythagorean theorem to find the magnitude of the resultant vector and select a suitable trigonometric function to find the angle that the resultant vector makes with the x axis. PITFALL PREVENTION 3.3 x and y Components Equations 3.8 and 3.9 associate the cosine of the angle with the x component and the sine of the angle with the y component. This is true only because we measured the angle ␪ with respect to the x axis, so don’t memorize these equations. If ␪ is measured with respect to the y axis (as in some problems), these equations will be incorrect. Think about which side of the triangle containing the components is adjacent to the angle and which side is opposite, and assign the cosine and sine accordingly. L PITFALL PREVENTION 3.4 Tangents on Calculators Generally, the inverse tangent function on calculators provides an angle between Ϫ 90° and ϩ 90°. As a consequence, if the vector you are studying lies in the second or third quadrant, the angle measured from the positive x axis will be the angle your calculator returns plus 180°.
    • C HAPTE R 3 • Vectors 68 Example 3.3 The Sum of Two Vectors Find the sum of two vectors A and B lying in the xy plane and given by A ϭ (2.0ˆ ϩ 2.0ˆ m i j) and B ϭ (2.0ˆ Ϫ 4.0ˆ m i j) Solution You may wish to draw the vectors to conceptualize the situation. We categorize this as a simple plug-in problem. Comparing this expression for A with the general expression A ϭ Ax ˆ ϩ Ay ˆ we see that Ax ϭ 2.0 m and Ay ϭ 2.0 m. i j, Likewise, Bx ϭ 2.0 m and By ϭ Ϫ 4.0 m. We obtain the resultant vector R, using Equation 3.14: R ϭ A ϩ B ϭ (2.0 ϩ 2.0)ˆ m ϩ (2.0 Ϫ 4.0)ˆ m i j ˆ Ϫ 2.0ˆ m ϭ (4.0 i j) or R x ϭ 4.0 m Example 3.4 R y ϭ Ϫ 2.0 m R ϭ √R x 2 ϩR y 2 ϭ √(4.0 m)2 ϩ (Ϫ2.0 m)2 ϭ √20 m ϭ 4.5 m We can find the direction of R from Equation 3.17: tan ␪ ϭ Ry Ϫ2.0 m ϭ ϭ Ϫ0.50 Rx 4.0 m Your calculator likely gives the answer Ϫ 27° for ␪ ϭ tanϪ1(Ϫ 0.50). This answer is correct if we interpret it to mean 27° clockwise from the x axis. Our standard form has been to quote the angles measured counterclockwise from the ϩ x axis, and that angle for this vector is ␪ ϭ 333° . The Resultant Displacement A particle undergoes three consecutive displacements: ˆ ˆ d1 ϭ (15ˆ ϩ 30ˆ ϩ 12 k) cm, d2 ϭ (23ˆ Ϫ 14ˆ Ϫ 5.0 k) cm i j i j ˆ ϩ 15ˆ cm. Find the components of the and d3 ϭ (Ϫ 13 i j) resultant displacement and its magnitude. Solution Three-dimensional displacements are more difficult to imagine than those in two dimensions, because the latter can be drawn on paper. For this problem, let us conceptualize that you start with your pencil at the origin of a piece of graph paper on which you have drawn x and y axes. Move your pencil 15 cm to the right along the x axis, then 30 cm upward along the y axis, and then 12 cm vertically away from the graph paper. This provides the displacement described by d1. From this point, move your pencil 23 cm to the right parallel to the x axis, 14 cm parallel to the graph paper in the Ϫ y direction, and then 5.0 cm vertically downward toward the graph paper. You are now at the displacement from the origin described by d1 ϩ d2. From this point, move your pencil 13 cm to the left in the Ϫ x direction, and (finally!) 15 cm parallel to the graph paper along the y axis. Example 3.5 The magnitude of R is found using Equation 3.16: Taking a Hike A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40.0 km in a direction 60.0° north of east, at which point she discovers a forest ranger’s tower. (A) Determine the components of the hiker’s displacement for each day. Solution We conceptualize the problem by drawing a sketch as in Figure 3.19. If we denote the displacement vectors on the first and second days by A and B, respectively, and use the car as the origin of coordinates, we obtain the vectors shown in Figure 3.19. Drawing the resultant R, we can now categorize this as a problem we’ve solved before—an addition of two vectors. This should give you a hint of the power of categorization— many new problems are very similar to problems that we have already solved if we are careful to conceptualize them. Your final position is at a displacement d1 ϩ d2 ϩ d3 from the origin. Despite the difficulty in conceptualizing in three dimensions, we can categorize this problem as a plug-in problem due to the careful bookkeeping methods that we have developed for vectors. The mathematical manipulation keeps track of this motion along the three perpendicular axes in an organized, compact way: R ϭ d1 ϩ d2 ϩ d3 ϭ (15 ϩ 23 Ϫ 13)ˆ cm ϩ (30 Ϫ 14 ϩ 15)ˆ cm i j ϩ (12 Ϫ 5.0 ϩ 0)ˆ cm k ˆ ϭ (25ˆ ϩ 31ˆ ϩ 7.0 k) cm i j The resultant displacement has components R x ϭ 25 cm, Ry ϭ 31 cm, and Rz ϭ 7.0 cm. Its magnitude is R ϭ √R x 2 ϩ R y 2 ϩ R z2 ϭ √(25 cm)2 ϩ (31 cm)2 ϩ (7.0 cm)2 ϭ 40 cm Interactive We will analyze this problem by using our new knowledge of vector components. Displacement A has a magnitude of 25.0 km and is directed 45.0° below the positive x axis. From Equations 3.8 and 3.9, its components are Ax ϭ A cos (Ϫ45.0Њ) ϭ (25.0 km)(0.707) ϭ 17.7 km Ay ϭ A sin(Ϫ45.0Њ) ϭ (25.0 km)(Ϫ0.707) ϭ Ϫ17.7 km The negative value of Ay indicates that the hiker walks in the negative y direction on the first day. The signs of Ax and Ay also are evident from Figure 3.19. The second displacement B has a magnitude of 40.0 km and is 60.0° north of east. Its components are
    • S ECTI O N 3.4 • Components of a Vector and Unit Vectors Bx ϭ B cos 60.0Њ ϭ (40.0 km)(0.500) ϭ 20.0 km By ϭ B sin 60.0Њ ϭ (40.0 km)(0.866) ϭ 34.6 km (B) Determine the components of the hiker’s resultant displacement R for the trip. Find an expression for R in terms of unit vectors. y(km) 0 Car –10 –20 R x ϭ Ax ϩ Bx ϭ 17.7 km ϩ 20.0 km ϭ 37.7 km R y ϭ Ay ϩ By ϭ Ϫ17.7 km ϩ 34.6 km ϭ 16.9 km R ϭ (37.7 ˆ ϩ 16.9 ˆ km i j) E S 20 Tower R 10 Solution The resultant displacement for the trip R ϭ A ϩ B has components given by Equation 3.15: In unit–vector form, we can write the total displacement as N W 45.0° 20 A 30 40 B 60.0° 50 x(km) Tent Figure 3.19 (Example 3.5) The total displacement of the hiker is the vector R ϭ A ϩ B. Using Equations 3.16 and 3.17, we find that the vector R has a magnitude of 41.3 km and is directed 24.1° north of east. Let us finalize. The units of R are km, which is reasonable for a displacement. Looking at the graphical representation in Figure 3.19, we estimate that the final position of the hiker is at about (38 km, 17 km) which is consistent with the components of R in our final result. Also, both components of R are positive, putting the final position in the first quadrant of the coordinate system, which is also consistent with Figure 3.19. Investigate this situation at the Interactive Worked Example link at http://www.pse6.com. Example 3.6 69 Let’s Fly Away! A commuter airplane takes the route shown in Figure 3.20. First, it flies from the origin of the coordinate system shown to city A, located 175 km in a direction 30.0° north of east. Next, it flies 153 km 20.0° west of north to city B. Finally, it flies 195 km due west to city C. Find the location of city C relative to the origin. Solution Once again, a drawing such as Figure 3.20 allows us to conceptualize the problem. It is convenient to choose the coordinate system shown in Figure 3.20, where the x axis points to the east and the y axis points to the north. Let us denote the three consecutive displacements by the vectors a, b, and c. We can now categorize this problem as being similar to Example 3.5 that we have already solved. There are two primary differences. First, we are adding three vectors instead of two. Second, Example 3.5 guided us by first asking for the components in part (A). The current Example has no such guidance and simply asks for a result. We need to analyze the situation and choose a path. We will follow the same pattern that we did in Example 3.5, beginning with finding the components of the three vectors a, b, and c. Displacement a has a magnitude of 175 km and the components a x ϭ a cos(30.0Њ) ϭ (175 km)(0.866) ϭ 152 km a y ϭ a sin(30.0Њ) ϭ (175 km)(0.500) ϭ 87.5 km Displacement b, whose magnitude is 153 km, has the components bx ϭ b cos(110Њ) ϭ (153 km)(Ϫ0.342) ϭ Ϫ52.3 km by ϭ b sin(110Њ) ϭ (153 km)(0.940) ϭ 144 km Finally, displacement c, whose magnitude is 195 km, has the components cx ϭ c cos(180Њ) ϭ (195 km)(Ϫ 1) ϭ Ϫ 195 km cy ϭ c sin(180Њ) ϭ 0 Therefore, the components of the position vector R from the starting point to city C are Rx ϭ ax ϩ bx ϩ cx ϭ 152 km Ϫ 52.3 km Ϫ 195 km ϭ Ϫ95.3 km Ry ϭ ay ϩ by ϩ cy ϭ 87.5 km ϩ 144 km ϩ 0 ϭ 232 km y(km) C N 250 B c W 200 S 20.0° 150 R b E 110° 100 50 A a 30.0° 50 100 150 200 x(km) Figure 3.20 (Example 3.6) The airplane starts at the origin, flies first to city A, then to city B, and finally to city C.
    • 70 C HAPTE R 3 • Vectors In unit–vector notation, R ϭ (Ϫ 95.3ˆ ϩ 232ˆ km . Using i j) Equations 3.16 and 3.17, we find that the vector R has a magnitude of 251 km and is directed 22.3° west of north. To finalize the problem, note that the airplane can reach city C from the starting point by first traveling 95.3 km due west and then by traveling 232 km due north. Or it could follow a straight-line path to C by flying a distance R ϭ 251 km in a direction 22.3° west of north. What If? After landing in city C, the pilot wishes to return to the origin along a single straight line. What are the components of the vector representing this displacement? What should the heading of the plane be? Take a practice test for this chapter by clicking on the Practice Test link at http://www.pse6.com. Answer The desired vector H (for Home!) is simply the negative of vector R: H ϭ Ϫ R ϭ (ϩ 95.3ˆ Ϫ 232ˆ km i j) The heading is found by calculating the angle that the vector makes with the x axis: tan ␪ ϭ Ry Ϫ232 m ϭ ϭ Ϫ2.43 Rx 95.3 m This gives a heading angle of ␪ ϭ Ϫ 67.7°, or 67.7° south of east. S U M MARY Scalar quantities are those that have only a numerical value and no associated direction. Vector quantities have both magnitude and direction and obey the laws of vector addition. The magnitude of a vector is always a positive number. When two or more vectors are added together, all of them must have the same units and all of them must be the same type of quantity. We can add two vectors A and B graphically. In this method (Fig. 3.6), the resultant vector R ϭ A ϩ B runs from the tail of A to the tip of B. A second method of adding vectors involves components of the vectors. The x component Ax of the vector A is equal to the projection of A along the x axis of a coordinate system, as shown in Figure 3.13, where Ax ϭ A cos ␪. The y component Ay of A is the projection of A along the y axis, where Ay ϭ A sin ␪. Be sure you can determine which trigonometric functions you should use in all situations, especially when ␪ is defined as something other than the counterclockwise angle from the positive x axis. If a vector A has an x component Ax and a y component Ay , the vector can be expressed in unit–vector form as A ϭ Ax ˆ ϩ Ay ˆ In this notation, ˆ is a unit vector pointi j. i ing in the positive x direction, and ˆ is a unit vector pointing in the positive y direction. j Because ˆ and ˆ are unit vectors, ͉ ˆ = ͉ ˆ = 1. i j i͉ j͉ We can find the resultant of two or more vectors by resolving all vectors into their x and y components, adding their resultant x and y components, and then using the Pythagorean theorem to find the magnitude of the resultant vector. We can find the angle that the resultant vector makes with respect to the x axis by using a suitable trigonometric function. QU ESTIONS 1. Two vectors have unequal magnitudes. Can their sum be zero? Explain. 2. Can the magnitude of a particle’s displacement be greater than the distance traveled? Explain. 4. Which of the following are vectors and which are not: force, temperature, the volume of water in a can, the ratings of a TV show, the height of a building, the velocity of a sports car, the age of the Universe? 3. The magnitudes of two vectors A and B are A ϭ 5 units and B ϭ 2 units. Find the largest and smallest values possible for the magnitude of the resultant vector R ϭ A ϩ B . 5. A vector A lies in the xy plane. For what orientations of A will both of its components be negative? For what orientations will its components have opposite signs?
    • Problems 6. A book is moved once around the perimeter of a tabletop with the dimensions 1.0 m ϫ 2.0 m. If the book ends up at its initial position, what is its displacement? What is the distance traveled? 7. While traveling along a straight interstate highway you notice that the mile marker reads 260. You travel until you reach mile marker 150 and then retrace your path to the mile marker 175. What is the magnitude of your resultant displacement from mile marker 260? 71 11. If A ϭ B, what can you conclude about the components of A and B? 12. Is it possible to add a vector quantity to a scalar quantity? Explain. 9. Can the magnitude of a vector have a negative value? Explain. 13. The resolution of vectors into components is equivalent to replacing the original vector with the sum of two vectors, whose sum is the same as the original vector. There are an infinite number of pairs of vectors that will satisfy this condition; we choose that pair with one vector parallel to the x axis and the second parallel to the y axis. What difficulties would be introduced by defining components relative to axes that are not perpendicular—for example, the x axis and a y axis oriented at 45° to the x axis? 10. Under what circumstances would a nonzero vector lying in the xy plane have components that are equal in magnitude? 14. In what circumstance is the x component of a vector given by the magnitude of the vector times the sine of its direction angle? 8. If the component of vector A along the direction of vector B is zero, what can you conclude about the two vectors? PROBLEMS 1, 2, 3 = straightforward, intermediate, challenging = full solution available in the Student Solutions Manual and Study Guide = coached solution with hints available at http://www.pse6.com = computer useful in solving problem = paired numerical and symbolic problems Section 3.1 1. Coordinate Systems on the opposite bank, she walks 100 m along the riverbank to establish a baseline. Then she sights across to the tree. The angle from her baseline to the tree is 35.0°. How wide is the river? The polar coordinates of a point are r ϭ 5.50 m and ␪ ϭ 240°. What are the Cartesian coordinates of this point? 2. Two points in a plane have polar coordinates (2.50 m, 30.0°) and (3.80 m, 120.0°). Determine (a) the Cartesian coordinates of these points and (b) the distance between them. 3. A fly lands on one wall of a room. The lower left-hand corner of the wall is selected as the origin of a two-dimensional Cartesian coordinate system. If the fly is located at the point having coordinates (2.00, 1.00) m, (a) how far is it from the corner of the room? (b) What is its location in polar coordinates? 4. Two points in the xy plane have Cartesian coordinates (2.00, Ϫ 4.00) m and (Ϫ 3.00, 3.00) m. Determine (a) the distance between these points and (b) their polar coordinates. 5. If the rectangular coordinates of a point are given by (2, y) and its polar coordinates are (r, 30°), determine y and r. 6. If the polar coordinates of the point (x, y) are (r, ␪), determine the polar coordinates for the points: (a) (Ϫ x, y ), (b) (Ϫ 2x, Ϫ2y), and (c) (3x, Ϫ 3y). Section 3.2 Vector and Scalar Quantities Section 3.3 Some Properties of Vectors 7. A surveyor measures the distance across a straight river by the following method: starting directly across from a tree 8. A pedestrian moves 6.00 km east and then 13.0 km north. Find the magnitude and direction of the resultant displacement vector using the graphical method. 9. A plane flies from base camp to lake A, 280 km away, in a direction of 20.0° north of east. After dropping off supplies it flies to lake B, which is 190 km at 30.0° west of north from lake A. Graphically determine the distance and direction from lake B to the base camp. 10. Vector A has a magnitude of 8.00 units and makes an angle of 45.0° with the positive x axis. Vector B also has a magnitude of 8.00 units and is directed along the negative x axis. Using graphical methods, find (a) the vector sum A ϩ B and (b) the vector difference A Ϫ B. 11. A skater glides along a circular path of radius 5.00 m. If he coasts around one half of the circle, find (a) the magnitude of the displacement vector and (b) how far the person skated. (c) What is the magnitude of the displacement if he skates all the way around the circle? 12. A force F1 of magnitude 6.00 units acts at the origin in a direction 30.0° above the positive x axis. A second force F2 of magnitude 5.00 units acts at the origin in the direction of the positive y axis. Find graphically the magnitude and direction of the resultant force F1 ϩ F2. 13. Arbitrarily define the “instantaneous vector height” of a person as the displacement vector from the point halfway
    • 72 C HAPTE R 3 • Vectors between his or her feet to the top of the head. Make an order-of-magnitude estimate of the total vector height of all the people in a city of population 100 000 (a) at 10 o’clock on a Tuesday morning, and (b) at 5 o’clock on a Saturday morning. Explain your reasoning. 14. A dog searching for a bone walks 3.50 m south, then runs 8.20 m at an angle 30.0° north of east, and finally walks 15.0 m west. Find the dog’s resultant displacement vector using graphical techniques. 15. Each of the displacement vectors A and B shown in Fig. P3.15 has a magnitude of 3.00 m. Find graphically (a) A ϩ B, (b) A Ϫ B, (c) B Ϫ A, (d) A Ϫ 2B. Report all angles counterclockwise from the positive x axis. 19. A vector has an x component of Ϫ 25.0 units and a y component of 40.0 units. Find the magnitude and direction of this vector. 20. A person walks 25.0° north of east for 3.10 km. How far would she have to walk due north and due east to arrive at the same location? 21. Obtain expressions in component form for the position vectors having the following polar coordinates: (a) 12.8 m, 150° (b) 3.30 cm, 60.0° (c) 22.0 in., 215°. 22. A displacement vector lying in the xy plane has a magnitude of 50.0 m and is directed at an angle of 120° to the positive x axis. What are the rectangular components of this vector? 23. A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. (a) What is her resultant displacement? (b) What is the total distance she travels? y B 3.00 m A 0m 3.0 30.0° O x Figure P3.15 Problems 15 and 37. 16. Three displacements are A ϭ 200 m, due south; B ϭ 250 m, due west; C ϭ 150 m, 30.0° east of north. Construct a separate diagram for each of the following possible ways of adding these vectors: R1 ϭ A ϩ B ϩ C; R2 ϭ B ϩ C ϩ A; R3 ϭ C ϩ B ϩ A. 17. A roller coaster car moves 200 ft horizontally, and then rises 135 ft at an angle of 30.0° above the horizontal. It then travels 135 ft at an angle of 40.0° downward. What is its displacement from its starting point? Use graphical techniques. Section 3.4 Components of a Vector and Unit Vectors 18. Find the horizontal and vertical components of the 100-m displacement of a superhero who flies from the top of a tall building following the path shown in Fig. P3.18. 24. In 1992, Akira Matsushima, from Japan, rode a unicycle across the United States, covering about 4 800 km in six weeks. Suppose that, during that trip, he had to find his way through a city with plenty of one-way streets. In the city center, Matsushima had to travel in sequence 280 m north, 220 m east, 360 m north, 300 m west, 120 m south, 60.0 m east, 40.0 m south, 90.0 m west (road construction) and then 70.0 m north. At that point, he stopped to rest. Meanwhile, a curious crow decided to fly the distance from his starting point to the rest location directly (“as the crow flies”). It took the crow 40.0 s to cover that distance. Assuming the velocity of the crow was constant, find its magnitude and direction. 25. While exploring a cave, a spelunker starts at the entrance and moves the following distances. She goes 75.0 m north, 250 m east, 125 m at an angle 30.0° north of east, and 150 m south. Find the resultant displacement from the cave entrance. 26. A map suggests that Atlanta is 730 miles in a direction of 5.00° north of east from Dallas. The same map shows that Chicago is 560 miles in a direction of 21.0° west of north from Atlanta. Modeling the Earth as flat, use this information to find the displacement from Dallas to Chicago. 27. Given the vectors A ϭ 2.00ˆ ϩ 6.00ˆ and B ϭ 3.00ˆ Ϫ i j i 2.00ˆ (a) draw the vector sum C ϭ A ϩ B and the vector j, difference D ϭ A Ϫ B. (b) Calculate C and D, first in terms of unit vectors and then in terms of polar coordinates, with angles measured with respect to the ϩ x axis. 28. Find the magnitude and direction of the resultant of three displacements having rectangular components (3.00, 2.00) m, (Ϫ 5.00, 3.00) m, and (6.00, 1.00) m. y 30.0° 100 m Figure P3.18 x 29. A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes an angle of 120° with the positive x axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.0° to the positive x axis. Find the magnitude and direction of the second displacement. 30. Vector A has x and y components of Ϫ 8.70 cm and 15.0 cm, respectively; vector B has x and y components of 13.2 cm and Ϫ 6.60 cm, respectively. If A Ϫ B ϩ 3C ϭ 0, what are the components of C?
    • Problems 31. Consider the two vectors A ϭ 3ˆ Ϫ 2ˆ and B ϭ Ϫ ˆ Ϫ 4ˆ i j i j. Calculate (a) A ϩ B, (b) A Ϫ B, (c) ͉A ϩ B͉, (d) ͉A Ϫ B͉, and (e) the directions of A ϩ B and A Ϫ B. 32. Consider the three displacement vectors A ϭ (3ˆ Ϫ 3ˆ m, i j) ˆ B ϭ (ˆ Ϫ 4ˆ m, and C ϭ (Ϫ 2i ϩ 5ˆ m. Use the compoi j) j) 73 through a quarter of a circle of radius 3.70 cm that lies in a north-south vertical plane. Find (a) the magnitude of the total displacement of the object, and (b) the angle the total displacement makes with the vertical. nent method to determine (a) the magnitude and direction of the vector D ϭ A ϩ B ϩ C, (b) the magnitude and direction of E ϭ Ϫ A Ϫ B ϩ C. 33. A particle undergoes the following consecutive displacements: 3.50 m south, 8.20 m northeast, and 15.0 m west. What is the resultant displacement? 34. In a game of American football, a quarterback takes the ball from the line of scrimmage, runs backward a distance of 10.0 yards, and then sideways parallel to the line of scrimmage for 15.0 yards. At this point, he throws a forward pass 50.0 yards straight downfield perpendicular to the line of scrimmage. What is the magnitude of the football’s resultant displacement? 35. The helicopter view in Fig. P3.35 shows two people pulling on a stubborn mule. Find (a) the single force that is equivalent to the two forces shown, and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons (abbreviated N). y F1 = 120 N F2 = 80.0 N 75.0˚ 60.0˚ x Figure P3.35 36. A novice golfer on the green takes three strokes to sink the ball. The successive displacements are 4.00 m to the north, 2.00 m northeast, and 1.00 m at 30.0° west of south. Starting at the same initial point, an expert golfer could make the hole in what single displacement? 37. Use the component method to add the vectors A and B shown in Figure P3.15. Express the resultant A ϩ B in unit–vector notation. 38. In an assembly operation illustrated in Figure P3.38, a robot moves an object first straight upward and then also to the east, around an arc forming one quarter of a circle of radius 4.80 cm that lies in an east-west vertical plane. The robot then moves the object upward and to the north, Figure P3.38 39. Vector B has x, y, and z components of 4.00, 6.00, and 3.00 units, respectively. Calculate the magnitude of B and the angles that B makes with the coordinate axes. 40. You are standing on the ground at the origin of a coordinate system. An airplane flies over you with constant velocity parallel to the x axis and at a fixed height of 7.60 ϫ 103 m. At time t ϭ 0 the airplane is directly above you, so that the vector leading from you to it is P0 ϭ (7.60 ϫ 103 m)ˆ At j. t ϭ 30.0 s the position vector leading from you to the airplane is P30 ϭ (8.04 ϫ 103 m)ˆ ϩ (7.60 ϫ 103 m)ˆ Dei j. termine the magnitude and orientation of the airplane’s position vector at t ϭ 45.0 s. 41. The vector A has x, y, and z components of 8.00, 12.0, and Ϫ 4.00 units, respectively. (a) Write a vector expression for A in unit–vector notation. (b) Obtain a unit–vector expression for a vector B one fourth the length of A pointing in the same direction as A. (c) Obtain a unit–vector expression for a vector C three times the length of A pointing in the direction opposite the direction of A. 42. Instructions for finding a buried treasure include the following: Go 75.0 paces at 240°, turn to 135° and walk 125 paces, then travel 100 paces at 160°. The angles are measured counterclockwise from an axis pointing to the east, the ϩ x direction. Determine the resultant displacement from the starting point. ˆ 43. Given the displacement vectors A ϭ (3ˆ Ϫ 4 ˆ ϩ 4 k) m and i j ˆ ϩ 3ˆ Ϫ 7 k) m, find the magnitudes of the vectors ˆ B ϭ (2 i j (a) C ϭ A ϩ B and (b) D ϭ 2A Ϫ B, also expressing each in terms of its rectangular components. 44. A radar station locates a sinking ship at range 17.3 km and bearing 136° clockwise from north. From the same station a rescue plane is at horizontal range 19.6 km, 153° clockwise from north, with elevation 2.20 km. (a) Write the position vector for the ship relative to the plane, letting ˆ i ˆ represent east, ˆ north, and k up. (b) How far apart are the j plane and ship?
    • 74 C HAPTE R 3 • Vectors 45. As it passes over Grand Bahama Island, the eye of a hurricane is moving in a direction 60.0° north of west with a speed of 41.0 km/h. Three hours later, the course of the hurricane suddenly shifts due north, and its speed slows to 25.0 km/h. How far from Grand Bahama is the eye 4.50 h after it passes over the island? 46. (a) Vector E has magnitude 17.0 cm and is directed 27.0° counterclockwise from the ϩ x axis. Express it in unit– vector notation. (b) Vector F has magnitude 17.0 cm and is directed 27.0° counterclockwise from the ϩ y axis. Express it in unit–vector notation. (c) Vector G has magnitude 17.0 cm and is directed 27.0° clockwise from the Ϫ y axis. Express it in unit–vector notation. 47. Vector A has a negative x component 3.00 units in length and a positive y component 2.00 units in length. (a) Determine an expression for A in unit–vector notation. (b) Determine the magnitude and direction of A. (c) What vector B when added to A gives a resultant vector with no x component and a negative y component 4.00 units in length? 52. Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A ϩ B to be larger than the magnitude of A Ϫ B by the factor n, what must be the angle between them? ˆ 53. A vector is given by R ϭ 2ˆ ϩ ˆ ϩ 3 k. Find (a) the magi j nitudes of the x, y, and z components, (b) the magnitude of R, and (c) the angles between R and the x, y, and z axes. 54. The biggest stuffed animal in the world is a snake 420 m long, constructed by Norwegian children. Suppose the snake is laid out in a park as shown in Figure P3.54, forming two straight sides of a 105° angle, with one side 240 m long. Olaf and Inge run a race they invent. Inge runs directly from the tail of the snake to its head and Olaf starts from the same place at the same time but runs along the snake. If both children run steadily at 12.0 km/h, Inge reaches the head of the snake how much earlier than Olaf? 48. An airplane starting from airport A flies 300 km east, then 350 km at 30.0° west of north, and then 150 km north to arrive finally at airport B. (a) The next