Chapter 5.3, Problem 28E

a.

To determine

To find : Velocity of the particle.

Answer to Problem 28E

The velocity of the function is $v\left(t\right)=6t\left(1-t\right)$ .

Explanation of Solution

Given information :

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

Calculation :

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

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

  $\begin{array}{l}x\left(t\right)=3t^2-2t^3\\ x\text{'}\left(t\right)=6t-6t^2\\ v\left(t\right)=6t\left(1-t\right)\end{array}$

Hence,

The velocity of the function is $v\left(t\right)=6t\left(1-t\right)$ .

b.

To determine

To find : Acceleration of the particle.

Answer to Problem 28E

The acceleration of the function is $a\left(t\right)=6-12t$ .

Explanation of Solution

Given information :

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

Calculation :

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

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

  $\begin{array}{l}x\left(t\right)=3t^2-2t^3\\ x\text{'}\left(t\right)=6t-6t^2\\ x\text{'}\text{'}\left(t\right)=6-12t\\ a\left(t\right)=6-12t\end{array}$

Hence,

The acceleration of the function is $a\left(t\right)=6-12t$ .

c.

To determine

To describe : The motion of the particle.

Answer to Problem 28E

The particle is moving to the right on interval $0\le x<1$ and to the left on the interval $1<x<\infty$ and at $t=0.5$ the particle stops being accelerated to the right and starts increasing accelerating to the left.

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 $v\left(t\right)=6t\left(1-t\right)$ and from part (b) acceleration is $a\left(t\right)=6-12t$ .

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:

  $\begin{array}{l}6t\left(1-t\right)=0\\ t\left(1-t\right)=0\\ t=0\text{ or }\left(1-t\right)=0\\ t=0\text{ or }1-t=0\\ t=0\text{ or }\left(1-t\right)=0\\ t=0\text{ or }t=1\end{array}$

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

Set $a\left(t\right)=0$ and solve for t:

  $\begin{array}{l}6-12t=0\\ 6=12t\\ \frac{6}{12}=t\\ 0.5=t\end{array}$

Therefore, at $t=0.5$ the particle stops being accelerated to the right and starts increasing accelerating to the left.

Hence,

The particle is moving to the right on interval $0\le x<1$ and to the left on the interval $1<x<\infty$ and at $t=0.5$ the particle stops being accelerated to the right and starts increasing accelerating to the left.