a. Answer The velocity is zero at t = 0 , t = 4 , t = 12 and t = 16 . Explanation Given information : The graph of the position function of a particle moving along a line is given below Calculation : Since, the slope of the tangent to the curve gives velocity. Now, if the slope is zero than the velocity is also zero. Now, from the given graph the slope is zero at approximately t = 0 , t = 4 , t = 12 and t = 16 . Therefore, the velocity is zero at approximately t = 0 , t = 4 , t = 12 and t = 16 . Hence, The velocity is zero at approximately t = 0 , t = 4 , t = 12 and t = 16 . b. To determine To find : the time at which the particle’s acceleration is zero. Answer The acceleration is zero at t = 1.5 , t = 5 , t = 8 , t = 11 and t = 13 . Explanation Given information : The graph of the position function of a particle moving along a line is given below. Calculation : The point where the concavity of the graph changes is the point where acceleration is zero. Since, the graph changes its concavity at t = 1.5 , t = 5 , t = 8 , t = 11 and t = 13 . Therefore, the acceleration is zero at t = 1.5 , t = 5 , t = 8 , t = 11 and t = 13 . Hence, The acceleration is zero at t = 1.5 , t = 5 , t = 8 , t = 11 and t = 13 .

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.3, Problem 30E

To determine

To find : the time at which the particle’s velocity is zero.

Expert Solution

Answer to Problem 30E

The velocity is zero at t=0 , t=4 , t=12 and t=16 .

Explanation of Solution

Given information :

The graph of the position function of a particle moving along a line is given below

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.3, Problem 30E , additional homework tip 1

Calculation :

Since, the slope of the tangent to the curve gives velocity. Now, if the slope is zero than the velocity is also zero.

Now, from the given graph the slope is zero at approximately t=0 , t=4 , t=12 and t=16 .

Therefore, the velocity is zero at approximately t=0 , t=4 , t=12 and t=16 .

Hence,

The velocity is zero at approximately t=0 , t=4 , t=12 and t=16 .

To determine

To find : the time at which the particle’s acceleration is zero.

Expert Solution

Answer to Problem 30E

The acceleration is zero at t=1.5 , t=5 , t=8 , t=11 and t=13 .

Explanation of Solution

Given information :

The graph of the position function of a particle moving along a line is given below.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.3, Problem 30E , additional homework tip 2

Calculation :

The point where the concavity of the graph changes is the point where acceleration is zero.

Since, the graph changes its concavity at t=1.5 , t=5 , t=8 , t=11 and t=13 .

Therefore, the acceleration is zero at t=1.5 , t=5 , t=8 , t=11 and t=13 .

Hence,

The acceleration is zero at t=1.5 , t=5 , t=8 , t=11 and t=13 .

Chapter 5.3, Problem 29E

Chapter 5.3, Problem 31E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

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