PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
PHYSICS FOR SCI.AND ENGR W/WEBASSIGN
10th Edition
ISBN: 9781337888462
Author: SERWAY
Publisher: CENGAGE L
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Textbook Question
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Chapter 2, Problem 10P

(a) Use the data in Problem 3 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 information, determine the average acceleration of the car. (d) What was the initial velocity of the car?

(a)

Expert Solution
Check Mark
To determine

The graph of position versus time.

Answer to Problem 10P

Therefore, the smooth graph of position versus time is shown in Figure I.

Explanation of Solution

The following table contains the data of position of the car at various time instants.

t(s)01.02.03.04.05.0
x(m)02.39.220.736.857.5

Draw the graph of position versus time for the derby car.

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 2, Problem 10P , additional homework tip  1

Figure I

In the shown graph, the position of the car at various time instants is plotted in the vertical axis against the time along horizontal direction.

Conclusion:

Therefore, the smooth graph of position versus time is shown in Figure I.

(b)

Expert Solution
Check Mark
To determine

The instantaneous velocity of the car at various time instants.

Answer to Problem 10P

The instantaneous velocity of the car at t=1s is 4.6m/s, at t=2s is 9.0m/s, at t=3s is 14m/s, at t=4s is 18m/s and at t=5.0s is 23m/s.

Explanation of Solution

The following table contains the data of position of the car at various time instants.

t(s)01.02.03.04.05.0
x(m)02.39.220.736.857.5

The instantaneous velocity is the slope of the tangent of the position versus time graph at an instant.

Formula to calculate the slope of the tangent is,

vt=1s=ΔxΔt                                                                    (I)

Here, vt=1s is the instantaneous velocity of the car or the slope of the tangent at t=1s, Δx is the position interval of the car and Δt is the time interval

Draw the tangent line at the time instant of t=1s in the graph of position versus time for the derby car.

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 2, Problem 10P , additional homework tip  2

Figure II

Substitute 4.6m for Δx and 1s for Δt in the above equation to find vt=1s.

    vt=1s=4.6m1s=4.6m/s

Therefore, the instantaneous velocity of the car at t=1s is 4.6m/s.

Draw the tangent line at the time instant of t=2s in the graph of position versus time for the derby car.

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 2, Problem 10P , additional homework tip  3

Figure III

Substitute 36m for Δx and 4.0s for Δt in the above equation to find vt=1s.

    vt=1s=36m4.0s=9.0m/s

Therefore, the instantaneous velocity of the car at t=2s is 9.0m/s.

Draw the tangent line at the time instant of t=3s in the graph of position versus time for the derby car.

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 2, Problem 10P , additional homework tip  4

Figure IV

Substitute 49m for Δx and 3.4s for Δt in the above equation to find vt=1s.

    vt=1s=49m3.4s=14m/s

Therefore, the instantaneous velocity of the car at t=3s is 14m/s.

Draw the tangent line at the time instant of t=4s in the graph of position versus time for the derby car.

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 2, Problem 10P , additional homework tip  5

Figure V

Substitute 54m for Δx and 3s for Δt in the above equation to find vt=1s.

    vt=1s=54m3s=18m/s

Therefore, the instantaneous velocity of the car at t=4s is 18m/s.

Draw the tangent line at the time instant of t=4s in the graph of position versus time for the derby car.

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 2, Problem 10P , additional homework tip  6

Figure VI

Substitute 58m for Δx and 2.5s for Δt in the above equation to find vt=1s.

    vt=1s=58m2.5s=23m/s

Conclusion:

Therefore, the instantaneous velocity of the car at t=1s is 4.6m/s, at t=2s is 9.0m/s, at t=3s is 14m/s, at t=4s is 18m/s and at t=5.0s is 23m/s.

(c)

Expert Solution
Check Mark
To determine

The average acceleration of the car.

Answer to Problem 10P

The average acceleration of the car is 2.3m/s2.

Explanation of Solution

The following table contains the instantaneous velocity of the car at various times instant.

t(s)1.02.03.04.05.0
v(t)m/s4.69.0141823

The graph of instantaneous velocity versus time for the derby car is shown below.

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN, Chapter 2, Problem 10P , additional homework tip  7

Figure VII

The figure VII shows that velocity of the car increases linearly, it means the acceleration of the car is constant throughout the motion.

Thus, the slope of this graph gives the average acceleration of the car.

Formula to calculate the slope of versus time graph is,

  a=ΔvΔt                                                                    (I)

Here, a is the average acceleration, Δv is the velocity interval and Δt is the time interval

Substitute 23m for Δx and 5.0s for Δt in the above equation to find vt=1s.

    vt=1s=23m/s5.0s=4.6m/s2

From the graph, the slope of the graph is 4.6m/s2.

Conclusion:

Therefore, the average acceleration of the car is 4.6m/s2.

(d)

Expert Solution
Check Mark
To determine

The initial velocity of the car.

Answer to Problem 10P

The initial velocity of the car is zero.

Explanation of Solution

The equation for the velocity of the car obtained from the graph is,

    v(t)=(2.3m/s2)t                          (I)

The first equation of motion gives the velocity of an object at any instant.

    v(t)=vi(t)+at                              (II)

Here, vi(t) is the initial velocity of the car.

Compare equation (I) and (II).

    vi(t)=0

Thus, the initial velocity of the car is zero.

Conclusion:

Therefore, the initial velocity of the car is zero.

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Chapter 2 Solutions

PHYSICS FOR SCI.AND ENGR W/WEBASSIGN

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