1. Chapter 4
ACCELERATION

2. change in velocity over the time interval over which it occurred
What is Acceleration?
Acceleration:
change in velocity over the time interval over which it occurred

3. Average Acceleration a = =
Average acceleration is the ratio of change in velocity over change in time.
v2 – v1 Δv
t2 – t Δt
a = =

4. Average Acceleration = = 4 m/s2 a = Example:
The velocity of a car increases from 2.0 m/s at 1.0 s to 16 m/s at 4.5 s. What is the car’s average acceleration?
16 m/s – 2.0 m/s m/s
4.5 s – 1. 0 s s =
= 4 m/s2
a =

5. Average Acceleration Example:
A car goes faster and faster backwards down a long driveway. We define forward velocity as positive, so backward velocity is negative. The car’s velocity changes from -2.0 m/s to -9.0 m/s in a 2.0-s time interval. Find the acceleration.

6. Average Acceleration Solution:
velocity change, Δv = -9.0 m/s – (-2.0 m/s)
= -7.0 m/s
average acceleration = Δv / Δt
= -7.0 m/s / 2 s
= -3.5 m / s^2

7. Average acceleration vs. Instantaneous acceleration
Average acceleration is calculated over a time interval.
Instantaneous acceleration occurs at a specific time.

8. Average Acceleration on Velocity-time graph
Slope of the line = average acceleration Δv Δt

9. Instantaneous Acceleration on Velocity-time graph
The slope of the tangent to the curve is the instantaneous acceleration for the object at the specified time, t. v t

10. Constant Acceleration
Acceleration that does not change in time is uniform or constant acceleration.
It is represented by a straight line on a velocity-time graph.

11. Constant Acceleration
vf = vi + at
Slope = a vi t

12. Constant Acceleration
Example:
If a car with a velocity of 2.0 m/s at t = 0 accelerates at a rate of +4.0 m/s^2 for 2.5 s, what is its velocity at time t = 2.5 s?
Using the equation of constant acceleration:
vf = vi + at
= +2.0 m/s + (+4.0 m/s2)(2.5 s)
= 12 m/s

13. Displacement when velocity and time are known
Average velocity: v = ½ (vf + vi)
Average velocity: v = d/t
Combining the two equations
d/t = ½ (vf + vi)
d = ½ (vf + vi) t

14. Displacement when acceleration and time are known
Starting with the displacement equation:
d = ½ (vf + vi) t
and using the final velocity equation:
vf = vi + at
Combining the two equations:
d = ½ (vi + at + vi) t = ½ (2vi + at) t
d = vit + ½ at2

15. Displacement when velocity and acceleration are known
Starting with d = ½ (vf + vi) t and
vf = vi + at
Substituting t = (vf – vi)/a into
d = ½ (vf + vi) t
d = ½ (vf + vi) (vf – vi) /a
= (vf 2 – vi2) / 2a
Solve for vf2:
vf 2 = vi2 + 2ad

16. Equations of Motion for Uniform Acceleration
Equations Variables
vf = vi + at vi vf a t
d = ½ (vf + vi) t vi vf d t
d = vit + ½ at2 vi a d t
vf 2 = vi 2 + 2ad vi vf a d

17. Example:
What is the displacement of a train as it is accelerated uniformly from +11 m/s to +33 m/s in a 20.0-s interval?

18. Example:
A car starting from rest accelerates uniformly at +6.1 m/s2 for 7.0 s. How far does the car move?

19. Example:
An airplane must reach a velocity of 71 m/s for takeoff. If the runway is 1.0 km long, what must the constant acceleration be?

20. Example:
The time the Demon Drop ride at Cedar Point, Ohio is freely falling for 1.5 s.
a. What is the velocity at the end of this time?
b. How far does it fall?

21. Problems
1. Find the uniform acceleration that causes a car’s velocity to change from 32 m/s to 96 m/s in an 8.0-s period.
96 m/s – 32 m/s = 8.0 m/s^2
8.0 s

22. Problems
2. Rocket-powered sleds are used to test the responses of humans to acceleration. Starting from rest, one sled can reach a speed of 444 m/s in 1.80 s and can be brought to a stop again in 2.15 s.

23. Problems
Calculate the acceleration of the sled when starting and compare it to the acceleration due to gravity, 9.80 m/s2.
a = 444 m/s – 0 = 247 m/s2
1.80 s
247 / 9.8 = 25.2

24. Problems
Find the acceleration of the sled when braking and compare it to the magnitude of the acceleration due to gravity.
a = 0 – 444 m/s = m/s2
2.15 s
207 m/s2 = 21.1
9.8 m/s2

25. Problems Find the acceleration of the moving object. 30 20 5 10 20 25
Velocity m/s 20 5 10 20 25
Time s

26. Problems During the first five seconds of travel.
a = 30 m/s – 0 m/s = 6 m/s2
5 s
b. Between the fifth and the tenth second of travel.
a. 30 m/s – 30 m/s = 0

27. Problems c. Between the tenth and the fifteenth second of travel.
a = 20 m/s – 30 m/s = -2 m/s2
5 s
d. Between the twentieth and the twenty-fifth second of travel.
a = 0 – 20 m/s = -4 m/s2

28. Problems
4. Position-time graph d d d t t t c a b

29. Problems
a. Velocity-time graphs v v v t t t b c a

30. Problems
b. Acceleration-time graphs a a a t t t b c a

31. Problems
5. A car with a velocity of 22 m/s is accelerated uniformly at the rate of 1.6 m/s^2 for 6.8 s. What is its final velocity?
vf = vi + at = 22 m/s + (1.6 m/s2)(6.8 s)
= 33 m/s

32. Problems
The velocity of an automobile changes over an 8.0-s time period as shown.
time velocity time velocity

33. Problems a. Plot the velocity-time graph of the motion. 20 16 12 8 4 2

34. Problems
b. Determine the displacement of the car the first 2.0 s. (displacement is the area under the curve)
d = ½ bh
= ½ (2.0 s)(8.0 m/s – 0)
= 8 m

35. Problems
c. What displacement does the car have during the first 4.0 s?
d = ½ bh
= ½ (4.0 s)(16.0 m/s – 0)
= 32 m

36. Problems
d. What displacement does the car have during the entire 8.0 s?
d = ½ bh + bh
= ½ (5.0 s)(20.0 m/s – 0)
+ (8.0 s – 5.0 s)(20.0 m/s)
= 110 m

37. Problems
e. Find the slope of the line between t = 0 s and t = 4.0 s. What does this slope represent?
a = Δv/ Δt
= (16 m/s – 0 m/s)/(4.0s – 0s)
= 4 m/s2

38. Problems
f. Find the slope of the line between t = 5.0 s and t = 7.0 s. What does this slope indicates.
a = Δv/ Δt
= (20 m/s – 20 m/s)/(7.0s – 5.0s)
= 0 constant velocity

39. Problems
7. Figure 4-20 shows the position-time and velocity-time graphs of a karate expert using a fist to break wooden boards.

40. Problems
7a. Use the velocity-time graph to describe the motion of the expert’s fist during the first 10 ms.
The fist moves downward at about -13 m/s for about 5 ms. It then suddenly comes to a halt (accelerates).

41. Problems
7b. Estimate the slope of the velocity-time graph to determine the acceleration of the fist when it suddenly stops.
a = Δv/ Δt
= (0 – (-13 m/s))/(7.5 ms – 5.0 ms))
= 5200 m/s2

42. Problems
7c. Express the acceleration as a multiple of the gravitational acceleration, g = 9.8 m/s^2.
(5200 m/s^2)/(9.8 m/s^2) = 530

43. Problems
7d. Determine the area under the velocity-time curve to find the displacement of the fist in the first 6 ms. Compare with the position-time graph.
The area is almost rectangular:
(-13 m/s)(0.006s) = -8 cm. This is in agreement with the position-time graph where the hand moves from +8 cm to 0 cm, for a net displacement of – 8cm.

44. Problems
8. A supersonic jet flying at 145 m/s is accelerated uniformly at the rate of 23.1 m/s^2 for 20 s.

45. Problems 8a. What is its final velocity? vf = vi + at
= 145 m/s + (23.1 m/s^2)(20.0 s)
= 607 m/s

46. Problems
b. The speed of sound in air is 331 m/s. How many times the speed of sound is the plane’s final speed?
x = (607 m/s) / (331 m/s)
= 1.83 times the speed of sound

47. Problems
Determine the final velocity of a proton that has an initial velocity of 2.35 x 10^5 m/s and then is accelerated uniformly in an electric field at the rate of x 10^12 m/s^2 for 1.50 x 10^-7 s.
vf = vi + at
= 235 m/s
+ (-1.10 x 10^12 m/s^2 )(1.50 x 10^-7 s)
= 7.0 x 10^4 m/s

48. Problems
10. Determine the displacement of a plane that is uniformly accelerated from 66 m/s to 88 m/s in 12 s.
d = (vf + vi)t / 2
= (88 m/s + 66 m/s)(12 s) / 2
= 924 m

49. Problems
11. How far does a plane fly in 15 s while its velocity is changing from +145 m/s to +75 m/s at a uniform rate of acceleration?
d = (vf + vi)t / 2
= (75 m/s m/s)(15 s) / 2
= 1650 m

50. Problems
12. A car moves at 12 m/s and coasts up a hill with a uniform acceleration of -1.6 m/s^2.

51. Problems How far has the car traveled after 6.0 s? d = vit + ½ at^2
= (12 m/s)(6 s) + ½ (-1.6 m/s)(6.0)^2
= 43 m
b. How far has it gone after 9.0 s?
= (12 m/s)(9 s) + ½ (-1.6 m/s)(9.0)^2

52. Problems
13. Four cars start from rest. Car A accelerates at 6.0 m/s^2, car b at 5.4 m/s^2, car C at 8.0 m/s^2, and car D at 12 m/s^2.

53. Problems 13 (a and b) Velocities of cars at the end of 2 s.
Car velocity (m/s) displacement (m) A B C D

54. Problems
13 c. What conclusions do you reach about the velocity attained and the displacement of a body starting from rest at the end of the first 2.0 s of acceleration?
The displacement traveled and velocity attained are numerically the same at the end of two seconds.

55. Problems
14. An astronaut drops a feather from 1.2 m above the surface of the moon. If the acceleration of gravity on the moon is 1.62 m/s^2, how long does it take the feather to hit the surface?

56. Problems 14. d = vit + ½ at^2 t = √(2d/g) = √((2)(-1.2 m)/(-1.6m/s2))

57. Problems
15. Table 4-4 is a table of displacement and velocities of a ball at the end of each second for the first 5.0 s of free fall from rest.
time displacement velocity

58. Problems 15 a. Use the data in the table to plot a velocity-time graph
Velocity
-50
5.0
Time

59. Problems
15 b. Use the data in the table to plot a position-time graph.
Position
-122.5
5.0
Time

60. Problems
15 c. Find the slope of the curve at the end of 2.0 and 4.0 s on the position-time graph. What are the approximate slopes? Do these values agree with the table of velocity?
at 2.0 s: slope = (-40 – (-1))/(3.0 – 1.0) =
-20 m/s
at 4.0 s: slope = (-118 – (-40))/(5.0 – 3.0) = -39 m/s
Yes, the values agree.

61. Problems
15d. Use the data in the table to plot a position versus time-squared graph. What time of curve is obtained? A straight line
-122.5 25
Time^2

62. Problems
15 e. Find the slope of the line at any point. Explain the significance of the value you obtain.
slope = ( – 0)/(25 – 0) = -4.9 m/s^2
The slope is ½ g

63. Problems 15 f. Does this curve agree with the equation d = ½ gt^2?
yes, since it is a straight line y = mx + b where y is d, m is ½ g, x is t^2 and b is 0.

64. Problems
16. A plane travels 5.0 x 10^2 m while being accelerated uniformly from rest at the rate of 5.0 m/s^2. What final velocity does it attain?
vf ^2 = vi^2 + 2ad
vf = √(vi^2 + 2ad)
= √(0^2+ 2(5.0)(500))
= 71 m/s

65. Problems
17. A race car can be slowed with a constant acceleration of -11m/s2.
If the car is going 55 m/s , how many meters will it take to stop?
Repeat for a car going 110 m/s.

66. 17a. vf2 = vi2 + 2ad
d = (vf2 – vi2)/2a
= (0 – 552)/(2*-11)
= 137 m

67. 17b. vf2 = vi2 + 2ad
d = (vf2 – vi2)/2a
= (0 – 1102)/(2*-11)
= 550 m

68. 25a. Determine if the car hits the barrier.
d = 50 meters, it hits the barrier
25b. V = 22 m/s

69. 26a.

70. 26b.
26c. Slope = s2/m
1/slope = 12 m/s2
26d. Yes, -6 m/s2

71. 28a. Dcar = 150 meters
28b. vf = 42 m/s

72. 29a.
29b. yes

73. 33a. vf = -15 m/s
33b. d = -10 m

74. 34a. vf = -25 m/s
34b. d = -30m
34c. The helicopter has fallen -10 m and the bag is 20 m below the helicopter
34d. The bag is 20 m below the helicopter after 2s.