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- 1. Kinematics in One Dimension
- 2. The Cheetah : A cat that is built for speed. Its strength and agility allow it to sustain a top speed of over 100 km/h. Such speeds can only be maintained for about ten seconds.
- 3. Objectives: After completing this lecture, you should be able to: <ul><li>Define and apply concepts of average and instantaneous velocity and acceleration. </li></ul><ul><li>Solve problems involving initial and final velocity , acceleration , displacement , and time . </li></ul><ul><li>Demonstrate your understanding of directions and signs for velocity, displacement, and acceleration. </li></ul><ul><li>Solve problems involving a free-falling body in a gravitational field . </li></ul>
- 4. Distance and Displacement <ul><li>Distance is the length of the actual path taken by an object. Consider travel from point A to point B in diagram below: </li></ul>Distance d is a scalar quantity (no direction): Contains magnitude only and consists of a number and a unit. A B d = 20 m
- 5. Distance and Displacement <ul><li>Displacement is the straight-line separation of two points in a specified direction. </li></ul>A vector quantity: Contains magnitude AND direction , a number, unit & angle. A B Δs = 12 m, 20 o
- 6. Distance and Displacement <ul><li>For motion along x or y axis, the displacement is determined by the x or y coordinate of its final position. Example: Consider a car that travels 8 m, E then 12 m, W. </li></ul>Net displacement Δx is from the origin to the final position: What is the distance traveled? d = 20 m !! Δx = 4 m, W x 12 m,W Δx 8 m,E x = +8 x = -4
- 7. The Signs of Displacement <ul><li>Displacement is positive (+) or negative (-) based on LOCATION . </li></ul>2 m -1 m -2 m The direction of motion does not matter! The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION. Examples:
- 8. Definition of Speed <ul><li>Speed is the distance traveled per unit of time (a scalar quantity). </li></ul>v s = 5 m/s Not direction dependent! d = 20 m Time t = 4 s v s = = d t 20 m 4 s A B
- 9. Definition of Velocity <ul><li>Velocity is the displacement per unit of time. (A vector quantity.) </li></ul>Direction required! A B s = 20 m Time t = 4 s Δx= 12 m 20 o = 3 m/s at 20 0 N of E
- 10. Average Speed and Instantaneous Velocity The instantaneous velocity is the magn-itude and direction of the velocity at a par-ticular instant. (v at point C) <ul><li>The average speed depends ONLY on the distance traveled and the time required. </li></ul>A B s = 20 m Time t = 4 s C
- 11. Example 1. A runner runs 200 m, east, then changes direction and runs 300 m, west . If the entire trip takes 60 s , what is the average speed and what is the average velocity? Recall that average speed is a function only of total distance and total time : Total distance: s = 200 m + 300 m = 500 m Direction does not matter! start Avg. speed= 8 m/s s 1 = 200 m s 2 = 300 m
- 12. Example 1 (Cont.) Now we find the average velocity, which is the net displacement divided by time . In this case, the direction matters. x o = 0 m; x = -100 m Direction of final displacement is to the left as shown. Note: Average velocity is directed to the west. x o = 0 t = 60 s x 1 = +200 m x = -100 m Average velocity:
- 13. Example 2. A sky diver jumps and falls for 600 m in 14 s. After chute opens, he falls another 400 m in 150 s. What is average speed for entire fall? Average speed is a function only of total distance traveled and the total time required. Total distance/ total time: 625 m 356 m 14 s 142 s A B
- 14. Examples of Speed Light = 3 x 10 8 m/s Orbit 2 x 10 4 m/s Jets = 300 m/s Car = 25 m/s
- 15. Speed Examples (Cont.) Runner = 10 m/s Snail = 0.001 m/s Glacier = 1 x 10 -5 m/s
- 16. The Signs of Velocity First choose + direction; then v is positive if motion is with that direction, and negative if it is against that direction. <ul><li>Velocity is positive (+) or negative (-) based on direction of motion. </li></ul>- + - + +
- 17. Definition of Acceleration <ul><li>An acceleration is the change in velocity per unit of time. (A vector quantity.) </li></ul><ul><li>A change in velocity requires the application of a push or pull ( force ). </li></ul>A formal treatment of force and acceleration will be given later. For now, you should know that: <ul><li>The direction of accel- eration is same as direction of force. </li></ul><ul><li>The acceleration is proportional to the magnitude of the force. </li></ul>
- 18. Acceleration and Force Pulling the wagon with twice the force produces twice the acceleration and acceleration is in direction of force. F a 2F 2a
- 19. Example of Acceleration The wind changes the speed of a boat from 2 m/s to 8 m/s in 3 s . Each second the speed changes by 2 m/s . Wind force is constant, thus acceleration is constant. + v = +8 m/s v 0 = +2 m/s t = 3 s Force
- 20. The Signs of Acceleration <ul><li>Acceleration is positive ( + ) or negative ( - ) based on the direction of force . </li></ul>Choose + direction first. Then acceleration a will have the same sign as that of the force F —regardless of the direction of velocity. a ( -) a (+) F F +
- 21. Example 3 (No change in direction): A constant force changes the speed of a car from 8 m/s to 20 m/s in 4 s . What is average acceleration? Step 1. Draw a rough sketch. Step 2. Choose a positive direction (right). Step 3. Label given info with + and - signs. Step 4. Indicate direction of force F. t = 4 s v = +20 m/s + v o = +8 m/s Force
- 22. Example 3 (Continued): What is average acceleration of car? Step 5. Recall definition of average acceleration. + v o = +8 m/s t = 4 s v = +20 m/s Force
- 23. Example 4: A wagon moving east at 20 m/s encounters a very strong head-wind, causing it to change directions. After 5 s , it is traveling west at 5 m/s . What is the average acceleration? (Be careful of signs.) Step 1. Draw a rough sketch. Step 2. Choose the eastward direction as positive. v o = +20 m/s v = -5 m/s Step 3. Label given info with + and - signs. + Force E
- 24. Example 4 (Cont.): Wagon moving east at 20 m/s encounters a head-wind, causing it to change directions. Five seconds later, it is traveling west at 5 m/s . What is the average acceleration? Choose the eastward direction as positive. Initial velocity, v o = +20 m/s, east (+) Final velocity, v f = -5 m/s, west (-) The change in velocity, v = v f - v 0 v = (-5 m/s) - (+20 m/s) = -25 m/s
- 25. Example 4: (Continued) = - 5 m/s 2 Acceleration is directed to left, west (same as F). = -25 m/s 5 s + Force v o = +20 m/s v = -5 m/s E v = (-5 m/s) - (+20 m/s) = -25 m/s
- 26. Signs for Displacement Time t = 0 at point A . What are the signs (+ or -) of displacement at B , C , and D ? At B, x is positive , right of origin At C , x is positive , right of origin At D , x is negative , left of origin + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D
- 27. Signs for Velocity What are the signs (+ or -) of velocity at points B, C, and D? <ul><li>At B, v is zero - no sign needed. </li></ul><ul><li>At C , v is positive on way out and negative on the way back. </li></ul><ul><li>At D , v is negative , moving to left. </li></ul>+ Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D x = 0
- 28. What are the signs (+ or -) of acceleration at points B, C, and D? <ul><li>The force is constant and always directed to left, so acceleration does not change. </li></ul><ul><li>At B, C, and D , a = -5 m/s, negative at all points. </li></ul>Signs for Acceleration + Force v o = +20 m/s v f = -5 m/s E a = - 5 m/s 2 A B C D
- 29. Definitions Average velocity: Average acceleration:
- 30. Graphical Analysis slope: velocity: acceleration:
- 31. x, (m) Position vs time graph (velocity)
- 32. v, (m/s) velocity vs time graph (acceleration)
- 33. Graphical Analysis Average Velocity: Instantaneous Velocity: x t x 2 x 1 t 2 t 1 x t Time slope Displacement, x
- 34. Uniform Acceleration in One Dimension: <ul><li>Motion is along a straight line (horizontal, vertical or slanted). </li></ul><ul><li>Changes in motion result from a CONSTANT force producing uniform acceleration. </li></ul><ul><li>The velocity of an object is changing by a constant amount in a given time interval. </li></ul><ul><li>The moving object is treated as though it were a point particle. </li></ul>
- 35. Average velocity for constant a : setting t o = 0 combining both equations: For constant acceleration:
- 36. Formulas based on definitions : Derived formulas : For constant acceleration only
- 37. Example 6: An airplane flying initially at 400 ft/s lands on a carrier deck and stops in a distance of 300 ft. What is the acceleration? Step 1. Draw and label sketch. Step 2. Indicate + direction Δ x = 300 ft v o = 400 ft/s v = 0 +
- 38. Example: (Cont.) Step 3. List given; find information with signs. Given: v o = 400 ft/s - initial velocity of airplane v = 0 - final velocity after traveling Δ x = +300 ft Find: a = ? - acceleration of airplane Δ x = 300 ft v o = 400 ft/s v = 0 +
- 39. Step 4. Select equation that contains a and not t . v 2 - v o 2 = 2 a Δ x a = - 300 ft/s 2 Why is the acceleration negative? Because Force is in a negative direction which means that the airplane slows down Given: v o = +400 ft/s v = 0 Δ x = +300 ft 0 a = = - v o 2 2x -(400 ft/s) 2 2(300 ft)
- 40. Example 5: A ball 5.0 m from the bottom of an incline is traveling initially at 8.0 m/s . Four seconds (4.0 s) later, it is traveling down the incline at 2.0 m/s . How far is it from the bottom at that instant? Given: d = 5.0 m - distance from initial position of the ball v o = 8.0 m/s - initial velocity v = -2.0 m/s - final velocity after t = 4.0 s Find: x = ? - distance from the bottom of the incline 5.0 m Δ x 8.0 m/s -2.0 m/s t = 4.0 s +
- 41. x = 17.0 m Given: d = 5.0 m v o = 8.0 m/s - initial velocity v = -2.0 m/s - final velocity after t = 4.0 s Find: x = ? - distance from the bottom of the incline Solution: where
- 42. Acceleration in our Example a = -2.50 m/s 2 What is the meaning of negative sign for a ? The force changing speed is down plane! 5 m x 8 m/s -2 m/s t = 4 s v o v + F
- 43. Use of Initial Position x 0 in Problems. If you choose the origin of your x,y axes at the point of the initial position, you can set x 0 = 0, simplifying these equations. The x o term is very useful for studying problems involving motion of two bodies. 0 0 0 0
- 44. Review of Symbols and Units <ul><li>Displacement ( x, x o ); meters ( m ) </li></ul><ul><li>Velocity ( v, v o ); meters per second ( m/s ) </li></ul><ul><li>Acceleration ( a ); meters per s 2 ( m/s 2 ) </li></ul><ul><li>Time ( t ); seconds ( s ) </li></ul>Review sign convention for each symbol
- 45. The Signs of Velocity <ul><li>Velocity is positive (+) or negative (-) based on direction of motion . </li></ul>First choose + direction; then velocity v is positive if motion is with that + direction, and negative if it is against that positive direction. + - - + +
- 46. The Signs of Displacement <ul><li>Displacement is positive (+) or negative (-) based on LOCATION . </li></ul>The displacement is the y-coordinate. Whether motion is up or down, + or - is based on LOCATION . 2 m -1 m -2 m
- 47. Acceleration Produced by Force <ul><li>Acceleration is ( + ) or ( - ) based on direction of force ( NOT based on v ). </li></ul>More will be said later on the relationship between F and a . A push or pull ( force ) is necessary to change velocity, thus the sign of a is same as sign of F . F a ( -) F a (+)
- 48. Problem Solving Strategy: <ul><li>Draw and label sketch of problem. </li></ul><ul><li>Indicate + direction and force direction. </li></ul><ul><li>List givens and state what is to be found. </li></ul>Given: ____, _____, _____ ( x,v,v o , a ,t ) Find: ____, _____ <ul><li>Select equation containing one and not the other of the unknown quantities, and solve for the unknown. </li></ul>
- 49. Acceleration Due to Gravity <ul><li>Every object on the earth experiences a common force: the force due to gravity. </li></ul><ul><li>This force is always directed toward the center of the earth (downward). </li></ul><ul><li>The acceleration due to gravity is relatively constant near the Earth’s surface. </li></ul>Earth W g
- 50. Gravitational Acceleration <ul><li>In a vacuum, all objects fall with same acceleration. </li></ul><ul><li>Equations for constant acceleration apply as usual. </li></ul><ul><li>Near the Earth’s surface: </li></ul>a = g = - 9.80 m/s 2 or -32 ft/s 2 Directed downward (usually negative).
- 51. Experimental Determination of Gravitational Acceleration. The apparatus consists of a device which measures the time required for a ball to fall a given distance. Suppose the height is 1.20 m and the drop time is recorded as 0.650 s. What is the acceleration due to gravity? y t
- 52. Experimental Determination of Gravity (y 0 = 0; y = -1.20 m) y = -1.20 m; t = 0.495 s Acceleration a is negative because force W is negative. y t Acceleration of Gravity: + W
- 53. Sign Convention: A Ball Thrown Vertically Upward <ul><li>Velocity is positive (+) or negative (-) based on direction of motion . </li></ul><ul><li>Displacement is positive (+) or negative (-) based on LOCATION . </li></ul>Release Point <ul><li>Acceleration is (+) or (-) based on direction of force (weight). </li></ul>y = 0 y = + y = + y = + y = 0 y = - Negative v = + v = 0 v = - v = - v= - Negative a = - a = - a = - a = - a = - UP = +
- 54. Same Problem Solving Strategy Except a = g : <ul><li>Draw and label sketch of problem. </li></ul><ul><li>Indicate + direction and force direction. </li></ul><ul><li>List givens and state what is to be found. </li></ul>Given: ____, _____, a = - 9.8 m/s 2 Find: ____, _____ <ul><li>Select equation containing one and not the other of the unknown quantities, and solve for the unknown. </li></ul>
- 55. Example 7: A ball is thrown vertically upward with an initial velocity of 30.0 m/s . What are its position and velocity after 2.00 s , 4.00 s , and 7.00 s ? Find also the maximum height attained v o = +30.0 m/s Given: a = - Δ 9.8 m/s 2 v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Find: Δ y = ? – displacement v = ? - final velocity After those three “times” Δ y = ? – maximum height a = g +
- 56. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Solutions: For t = 2.00 s: For t = 4.00 s: For t = 7.00 s:
- 57. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Solutions: For t = 2.00 s: For t = 4.00 s: For t = 7.00 s:
- 58. Given: a = -9.8 m/s 2 ; v o = 30.0 m/s t = 2.00 s; 4.00 s; 7.00 s Solutions: For maximum height, v = 0 (the ball stops at maximum height):
- 59. Average and Instantaneous a v t v 2 v 1 t 2 t 1 v t time slope
- 60. Experiment 10 Uniformly Accelerated Motion (Acceleration due to Gravity) 39 (06A)
- 61. Summary of Formulas Derived Formulas : For Constant Acceleration Only
- 62. Summary: Procedure <ul><li>Draw and label sketch of problem. </li></ul><ul><li>Indicate + direction and force direction. </li></ul><ul><li>List givens and state what is to be found. </li></ul>Given: ____, _____, ______ Find: ____, _____ <ul><li>Select equation containing one and not the other of the unknown quantities, and solve for the unknown. </li></ul>

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