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Chapter 2


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  • 1. Motion in One Dimension
    Chapter 2
    Pg. 40-65
  • 2. 2.1 Displacement and Velocity
  • 3. What do you think?
    Is the book on your instructor’s desk in motion?
    Explain your answer.
    What are some examples of motion?
    How would you define motion?
  • 4. One-Dimensional Motion
    It is the simplest form of motion
    This is motion that happens in one direction
    Train on the tracts moving forward or backward
  • 5. Displacement (x)
    Change in position
    Can be positive or negative
    It describes the direction of motion
    You will see me use d instead of xfor displacement
  • 6. Displacement (x)
    Displacement is not always equal to the distance traveled
    A gecko runs up a tree from the 20 cm marker to the 80 cm marker, then he retreats to the 50 cm marker?
  • 7. Displacement (x)
    In total it traveled 90 cm, however, the displacement is only 30 cm
    x or d= 50cm- 20cm = 30 cm
    If the gecko goes back to the start
    The displacement is zero
  • 8. Displacement
    What is the displacement for the objects shown?
    Answer: 70 cm
    • Answer: -60 cm
  • Displacement - Sign Conventions
    • Right (or east) ---> +
    Left (or west) ---> –
    Up (or north) ----> +
    Down (or south) ---> –
  • 9. Average Velocity
    • Average velocity is displacement divided by the time interval.
    The total displacement divided by the time interval during which the displacement occurred (Vavg)
    Basically velocity is distance/time
  • 10. Average Velocity
    The units can be determined from the equation.
    SI Units: m/s
    Other Possible Units: mi/h, km/h, cm/year
    Velocity can be positive or negative but time must always be positive
  • 11. Average Velocity
    You travel 370 km west to a friends house. You left at 10am and arrived at 3pm. What was your average velocity?
    Vavg = d = -370 km
    Δt 5.0h
    Vavg= -74 km/h or 74 km/h west
  • 12. Classroom Practice Problems
    Joey rides his bike for 15 min with an average velocity of 12.5 km/h, how far did he ride?
    Vavg = d
    d = (12.5 km/h) (0.25h)
    d = 3.125 km
  • 13. Speed
    Speed does not include direction while velocity does.
    Speed uses distance rather than displacement.
    In a round trip, the average velocity is zero but the average speed is not zero.
  • 14. Graphing Motion
    How would you describe the motion shown by this graph?
    Answer: Constant speed (straight line)
    What is the slope of this line?
    Answer: 1 m/s
    What is the average velocity?
    Answer: 1 m/s
  • 15. Graphing Motion
    Describe the motion of each object.
    Object 1: constant velocity to the right or upward
    Object 2: constant velocity of zero (at rest)
    Object 3: constant velocity to the left or downward
  • 16. 2.2 Acceleration
    Pg. 48 to 59
  • 17. What do you think?
    Which of the following cars is accelerating?
    A car shortly after a stoplight turns green
    A car approaching a red light
    A car with the cruise control set at 80 km/h
    A car turning a curve at a constant speed
    Based on your answers, what is your definition of acceleration?
  • 18. Acceleration
    What are the units?
    SI Units: m/s2
    Other Units: (km/h)/s or (mi/h)/s
    Acceleration = 0 implies a constant velocity (or rest)
  • 19. Classroom Practice Problem
    A bus slows down with an average acceleration of -1.8m/s2. how long does it take the bus to slow from 9.0m/s to a complete stop?
    Find the acceleration of an amusement park ride that falls from rest to a velocity of 28 m/s downward in 3.0 s.
    9.3 m/s2 downward
  • 20. Acceleration
    Displacement with constant acceleration
    d = ½ (vi + vf) Δt
    Displacement= ½ (initial velocity + final velocity) (time)
  • 21. Example
    A racecar reaches a speed of 42m/s. it uses its parachute and breaks to stop 5.5s later. Find the distance that the car travels during breaking.
    Given: vi = 42 m/s vf= 0 m/s
    t = 5.5 sec d = ??
  • 22. Example
    d = ½ (vi + vf) Δt
    d = ½ (42+0) (5.5)
    d = 115.5 m
  • 23. Acceleration
    Final velocity
    Depends on initial velocity, acceleration and time
    Velocity with constant acceleration
    vf = vi + aΔt
    Final velocity = initial velocity + (acceleration X time)
  • 24. Acceleration
    Displacement with constant acceleration
    d = viΔt + ½ a (Δt)2
    Displacement = (initial velocity X time) + ½ acceleration X time2
  • 25. Example
    A plane starting at rest at one end of a runway undergoes uniform acceleration of 4.8 m/s2 for 15s before takeoff. What is its speed at takeoff? How long must the runway before the plane to be able to take off?
    Given: vi = 0 m/s a= 4.8m/ss
    t= 15s vf = ?? d= ??
  • 26. Example
    What do we need to figure out?
    Speed at takeoff
    How long the runway needs to be
    vf = vi + aΔt
    vf = 0m/s + (4.8 m/s2)(15 s)
    vf = 72m/s
  • 27. Example
    d = viΔt + ½ a (Δt)2
    d = (0m/s)(15s) + ½ (4.8m/s2)(15s)2
    d = 540 m
  • 28. Acceleration
    Final velocity after any displacement
    vf2 = vi2 +2ad
    Final velocity2 = initial velocity2 + 2(acceleration)(displacement)
  • 29. Example
    An aircraft has a landing speed of 83.9 m/s. The landing area of an aircraft carrier is 195 m long. What is the minimum uniform acceleration required for safe landing?
    Answer: -18.0 m/s2
  • 30. Useful Equations
    d = 1 (vi + vf) t
  • 31. Classroom Practice Problems
    A bicyclist accelerates from 5.0 m/s to 16 m/s in 8.0 s. Assuming uniform acceleration, what distance does the bicyclist travel during this time interval?
    Answer: 84 m
  • 32. 2.3 Falling Objects
  • 33. Free Fall
    The motion of a body when only the force due to gravity is acting
    Acceleration is constant for the entire fall
    Acceleration due to gravity (ag or g)
    Has a value of -9.81 m/s2 (we will use 10m/s2)
    Negative for downward
    Roughly equivalent to -22 (mi/h)/s
  • 34. Free Fall
    Acceleration is constant during upward and downward motion
    Objects thrown into the air have a downward acceleration as soon as they are released.
  • 35. Free Fall
    The instant the velocity of the ball is equal to 0 m/s is the instant the ball reaches the peak of its upward motion and is about to begin moving downward.
    Although the velocity is 0m/s the acceleration is still equal to 10 m/s2
  • 36. Example
    Jason hits a volleyball so that it moves with an initial velocity of 6m/s straight upward. If the volleyball starts from 2.0m above the floor how long will it be in the air before it strikes the floor?
    We want to use our velocity and acceleration equations
  • 37. Example
    Given: vi = 6m/s a= 10m/s2
    d= -2.0m t = ?? vf = ??
    vf2= vi2 + 2ad
    vf= at + vi
  • 38. Free Fall
    For a ball tossed upward, make predictions for the sign of the velocity and acceleration to complete the chart.
  • 39. Graphing Free Fall
    Based on your present understanding of free fall, sketch a velocity-time graph for a ball that is tossed upward (assuming no air resistance).
    Is it a straight line?
    If so, what is the slope?
    Compare your predictions to the graph to the right.
  • 40. Classroom Practice Problem
    A ball is thrown straight up into the air at an initial velocity of 25.0 m/s upward. Create a table showing the ball’s position, velocity and acceleration each second for the first 5 s.
  • 41. Direction of Acceleration
    Describe the motion of an object with
    vi and a as shown to the left.
    Moving right as it speeds up
    Moving right as it slows down
    Moving left as it speeds up
    Moving left as it slows down
  • 42. Graphing Velocity
    The slope (rise/run) of a velocity/time graph is the acceleration.
    Rise is change in v
    Run is change in t
    This graph shows a constant acceleration.
    Average speed is the midpoint.
  • 43. Graph of v vs. t for a train
    Describe the motion at points A, B, and C.
    A: accelerating (increasing velocity/slope) to the right
    B: constant velocity to the right
    C: negative acceleration (decreasing velocity/slope) and still moving to the right