Chapter 5.3, Problem 25E

a.

To determine

To find : Velocity of the particle.

Answer to Problem 25E

The velocity of the function is $v\left(t\right)=2t-4$ .

Explanation of Solution

Given information :

The position function of the particle is $x\left(t\right)=t^2-4t+3$ .

Calculation :

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

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

  $\begin{array}{l}x\left(t\right)=t^2-4t+3\\ x\text{'}\left(t\right)=2t-4\\ v\left(t\right)=2t-4\end{array}$

Hence,

The velocity of the function is $v\left(t\right)=2t-4$ .

b.

To determine

To find : Acceleration of the particle.

Answer to Problem 25E

The acceleration of the function is $a\left(t\right)=2$ .

Explanation of Solution

Given information :

The position function of the particle is $x\left(t\right)=t^2-4t+3$ .

Calculation :

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

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

  $\begin{array}{l}x\left(t\right)=t^2-4t+3\\ x\text{'}\left(t\right)=2t-4\\ x\text{'}\text{'}\left(t\right)=2\\ a\left(t\right)=2\end{array}$

Hence,

The acceleration of the function is $a\left(t\right)=2$ .

c.

To determine

To describe : The motion of the particle.

Answer to Problem 25E

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

Explanation of Solution

Given information :

The position function of the particle is $x\left(t\right)=t^2-4t+3$ .

Calculation :

From part (a) velocity is $v\left(t\right)=2t-4$ and from part (b) acceleration is $a\left(t\right)=2$ .

When the function $x\left(t\right)$ 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 $v\left(t\right)=0$ and solve for t:

  $\begin{array}{l}2t-4=0\\ 2t=4\\ t=2\end{array}$

The zero partitions the t axis into two intervals as shown in the sign graph of v below:

  

The particle is moving to the right on interval $2<x<\infty$ and to the left on the interval $0\le x<2$ .

And the acceleration is constant throughout.

Hence,

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