Chapter 5.3, Problem 27E

a.

To determine

To find : Velocity of the particle.

Answer to Problem 27E

The velocity of the function is $v\left(t\right)=3t^2-3$ .

Explanation of Solution

Given information :

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

Calculation :

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

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

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

Hence,

The velocity of the function is $v\left(t\right)=3t^2-3$ .

b.

To determine

To find : Acceleration of the particle.

Answer to Problem 27E

The acceleration of the function is $a\left(t\right)=6t$ .

Explanation of Solution

Given information :

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

Calculation :

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

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

  $\begin{array}{l}x\left(t\right)=t^3-3t+3\\ x\text{'}\left(t\right)=3t^2-3\\ x\text{'}\text{'}\left(t\right)=6t\\ a\left(t\right)=6t\end{array}$

Hence,

The acceleration of the function is $a\left(t\right)=6t$ .

c.

To determine

To describe : The motion of the particle.

Answer to Problem 27E

The particle is moving to the right on interval $1<x<\infty$ and to the left on the interval $0\le x<1$ and the acceleration is towards the right for $t>0$ .

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 $v\left(t\right)=3t^2-3$ and from part (b) acceleration is $a\left(t\right)=6t$ .

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:

  $\begin{array}{l}3t^2-3=0\\ 3\left(t^2-1\right)=0\\ t^2-1=0\\ \left(t-1\right)\left(t+1\right)=0\\ t-1=0\text{ or }t+1=0\\ t=1\text{ or }t=-1\end{array}$

Since, $t\ne -1$ , therefore, $t=1$ .

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 $1<x<\infty$ and to the left on the interval $0\le x<1$ .

Since, there is no acceleration for $t=0$ . Therefore, the acceleration is towards the right for $t>0$ .

Hence,

The particle is moving to the right on interval $1<x<\infty$ and to the left on the interval $0\le x<1$ and the acceleration is towards the right for $t>0$ .