Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)
Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)
4th Edition
ISBN: 9780133178579
Author: Ross L. Finney
Publisher: PEARSON
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Chapter 5.3, Problem 26E

a.

To determine

To find : Velocity of the particle.

a.

Expert Solution
Check Mark

Answer to Problem 26E

The velocity of the function is v(t)=22t .

Explanation of Solution

Given information :

The position function of the particle is x(t)=62tt2 .

Calculation :

Since, the first derivative of the position function gives velocity. Therefore,

Find the first derivative to get the velocity of the particle:

  x(t)=62tt2x'(t)=22tv(t)=22t

Hence,

The velocity of the function is v(t)=22t .

b.

To determine

To find : Acceleration of the particle.

b.

Expert Solution
Check Mark

Answer to Problem 26E

The acceleration of the function is a(t)=2 .

Explanation of Solution

Given information :

The position function of the particle is x(t)=62tt2 .

Calculation :

Since, the second derivative of the position function gives acceleration. Therefore,

Find the second derivative to get the velocity of the particle:

  x(t)=62tt2x'(t)=22tx''(t)=2a(t)=2

Hence,

The acceleration of the function is a(t)=2 .

c.

To determine

To describe : The motion of the particle.

c.

Expert Solution
Check Mark

Answer to Problem 26E

The particle is moving to the right on interval 2<x< and to the left on the interval 0x<2 . And the acceleration is constant throughout.

Explanation of Solution

Given information :

The position function of the particle is x(t)=t24t+3 .

Calculation :

From part (a) velocity is v(t)=22t and from part (b) acceleration is a(t)=2 .

When the function x(t) is increasing the particle moves to the right to the x axis and when the function is decreasing the particle move to the left of the x axis.

Set v(t)=0 and solve for t:

  22t=02t=2t=1

Since, t is negative. Therefore, x(t) is decreasing.

The particle is moving left on the interval 0t< .

And the acceleration is constant throughout, accelerating the particle to the left.

Hence,

The particle is moving left on the interval 0t< . And the acceleration is constant throughout, accelerating the particle to the left.

Chapter 5 Solutions

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

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