Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 2, Problem 62AP

(a)

To determine

Acceleration of object from 0s to 4s.

(a)

Expert Solution
Check Mark

Answer to Problem 62AP

Acceleration of object from 0s to 4s is 0m/s2

Explanation of Solution

Write the equation for acceleration.

    `a=ΔvΔt

Here, a is the acceleration, Δv is the change in velocity, and Δt is the time interval.

The straight line parallel to x-axis of velocity-time graph from 0s to 4s shows that the velocity is constant.

Conclusion:

Substitute 0m/s for Δv in the above equation to find a.

    a=0m/sΔt=0m/s2

Therefore, the acceleration of object from 0s to 4s is 0m/s2

(b)

To determine

Acceleration of object from 4s to 9s.

(b)

Expert Solution
Check Mark

Answer to Problem 62AP

Acceleration of object from 4s to 9s is 6.0m/s2.

Explanation of Solution

Write the equation for acceleration.

    `a=ΔvΔt

Write the equation for Δv.

    Δv=vfvi

Here, vf is the final velocity and vi is the initial velocity.

Write the equation for Δt.

    Δt=tfti

Here, tf is the final time and ti is the initial time.

Rewrite the equation for a by substituting the above relations.

    a=vfvitfti                                                                                                          (I)

Conclusion:

Substitute 18m/s for vf, 12m/s for vi, 9s for tf, and 4s for ti in the above equation to find a.

    a=18m/s(12m/s)9s4s=6.0m/s2

Therefore, the acceleration of object from 4s to 9s is 6.0m/s2.

(c)

To determine

Acceleration of object from 13.0s to 18.0s.

(c)

Expert Solution
Check Mark

Answer to Problem 62AP

Acceleration of object from 13.0s to 18.0s is 3.6m/s2.

Explanation of Solution

Substitute 0m/s for vf, 18m/s for vi, 18s for tf, and 13s for ti in the above equation to find a.

    a=0m/s18m/s18s13s=3.6m/s2

Therefore, the acceleration of object from 13.0s to 18.0s is 3.6m/s2.

(d)

To determine

The time interval(s) at which the object moves with lowest speed.

(d)

Expert Solution
Check Mark

Answer to Problem 62AP

Object moves with lowest speed at 6s and at 18s.

Explanation of Solution

Speed will be either zero or positive only. In the graph , it can be seen that vx is zero at 6s and at 18s.

Therefore, the object moves with lowest speed at 6s and at 18s.

(e)

To determine

The time at which object is at maximum distance from x=0.

(e)

Expert Solution
Check Mark

Answer to Problem 62AP

The time at which object is at maximum distance from x=0 at t=18s.

Explanation of Solution

Initially, the object will goes into negative coordinates. At t=6s, object changes its direction. Up to t=18s, the object continues to move in same direction. So it can be say that the object is at maximum distance at t=18s.

Therefore, the time at which object is at maximum distance from x=0 at t=18s.

(f)

To determine

The final position of object at t=18s.

(f)

Expert Solution
Check Mark

Answer to Problem 62AP

The final position of object at t=18s will be x=84m.

Explanation of Solution

The distance travelled from velocity-time graph can be found using the area under the graph. The whole area of graph can be divided to five parts.

Write the equation to find the position of object at t=18s.

    Δx=a1b1+12a1b2+12a2b3+a2b4+12a2b5

Here, Δx is the final position of object at t=18s, a1anda2 are lengths in y-coordinates, and b1,b2,b3,b4,b5 are the lengths along x-axis.

Conclusion:

Substitute 12m/s for a1,18m/s for a2 4s for b1, 2s for b2, 3s for b3, 4s for b4, and 5s for b5 in the above equation to find Δx.

    Δx=(12m/s)(4s)+12(12m/s)(2s)+12(18m/s)(3s)+(18m/s)(4s)+12(18m/s)(5s)=48m12m+27m+72m+45m=84m

Therefore, the final position of object at t=18s will be x=84m.

(g)

To determine

The total distance travelled by object during the time from x=0s to t=18s.

(g)

Expert Solution
Check Mark

Answer to Problem 62AP

The total distance travelled by object during the time from x=0s to t=18s is 204m.

Explanation of Solution

Write the equation to find the total distance travelled by object.

    x=|a1b1|+|12a1b2|+|12a2b3|+|a2b4|+|12a2b5|

Here, x is the total distance travelled.

Conclusion:

Substitute 12m/s for a1,18m/s for a2 4s for b1, 2s for b2, 3s for b3, 4s for b4, and 5s for b5 in the above equation to find Δx.

    Δx=|(12m/s)(4s)|+|12(12m/s)(2s)|+|12(18m/s)(3s)|+|(18m/s)(4s)|+|12(18m/s)(5s)|=|48m|+|12m|+|27m|+|72m|+|45m|=204m

Therefore, the total distance travelled by object during the time from x=0s to t=18s is 204m.

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Chapter 2 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

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