a. Answer The velocity of the function is v ( t ) = − 2 − 2 t . Explanation Given information : The position function of the particle is x ( t ) = 6 − 2 t − t 2 . 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 ) = 6 − 2 t − t 2 x ' ( t ) = − 2 − 2 t v ( t ) = − 2 − 2 t Hence, The velocity of the function is v ( t ) = − 2 − 2 t . b. To determine To find : Acceleration of the particle. Answer The acceleration of the function is a ( t ) = − 2 . Explanation Given information : The position function of the particle is x ( t ) = 6 − 2 t − t 2 . 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 ) = 6 − 2 t − t 2 x ' ( t ) = − 2 − 2 t x ' ' ( t ) = − 2 a ( t ) = − 2 Hence, The acceleration of the function is a ( t ) = − 2 . c. To determine To describe : The motion of the particle. Answer The particle is moving to the right on interval 2 < x < ∞ and to the left on the interval 0 ≤ x < 2 . And the acceleration is constant throughout. Explanation Given information : The position function of the particle is x ( t ) = t 2 − 4 t + 3 . Calculation : From part (a) velocity is v ( t ) = − 2 − 2 t 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: − 2 − 2 t = 0 − 2 t = 2 t = − 1 Since, t is negative. Therefore, x ( t ) is decreasing. The particle is moving left on the interval 0 ≤ t < ∞ . And the acceleration is constant throughout, accelerating the particle to the left. Hence, The particle is moving left on the interval 0 ≤ t < ∞ . And the acceleration is constant throughout, accelerating the particle to the left.

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 26E

To determine

To find : Velocity of the particle.

Expert Solution

Answer to Problem 26E

The velocity of the function is v(t)=−2−2t .

Explanation of Solution

Given information :

The position function of the particle is x(t)=6−2t−t2 .

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)=6−2t−t2x'(t)=−2−2tv(t)=−2−2t

Hence,

The velocity of the function is v(t)=−2−2t .

To determine

To find : Acceleration of the particle.

Expert Solution

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)=6−2t−t2 .

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)=6−2t−t2x'(t)=−2−2tx''(t)=−2a(t)=−2

Hence,

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

To determine

To describe : The motion of the particle.

Expert Solution

Answer to Problem 26E

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

Explanation of Solution

Given information :

The position function of the particle is x(t)=t2−4t+3 .

Calculation :

From part (a) velocity is v(t)=−2−2t 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:

−2−2t=0−2t=2t=−1

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

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

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

Hence,

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

Chapter 5.3, Problem 25E

Chapter 5.3, Problem 27E

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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