Chapter 8.1, Problem 6E

a.

To determine

To determine the time when the particle is moving to the right , to the left, and stopped.

Answer to Problem 6E

The particle will move right when $0\le t<4$ .

The particle is not moving left.

The particle will stop when $t=4$ .

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=\sqrt{4-t},0\le t\le 4$

Calculation:

For the particle to move right, the velocity of particle must be greater than $0$ .

Therefore $v\left(t\right)>0$

  $\begin{array}{l}v\left(t\right)>0\\ \sqrt{4-t}>0\\ 4-t>0\\ t-4<0\\ \because 0\le t\le 4\\ \therefore 0\le t<4\end{array}$

For the particle to move left, the velocity of particle must be less than $0$ ,

Therefore $v\left(t\right)<0$

  $\begin{array}{l}v\left(t\right)<0\\ \sqrt{4-t}<0\\ \text{this statement is not true because there is no possible values of }t\\ \therefore \text{particle is not moving left}\end{array}$

For the particle to stop , its velocity must be equal to $0$

Therefore, $v\left(t\right)=0$

  $\begin{array}{l}v\left(t\right)=0\\ \sqrt{4-t}=0\\ 4-t=0\\ t=4\end{array}$

b.

To determine

To find the particle’s displacement for the time interval $0\le t\le 4$ and final position of particle if $s\left(0\right)=3$ .

Answer to Problem 6E

The displacement of particle is $5.34m$ and the final position of the particle is $8.34m$ from initial position.

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=\sqrt{4-t},0\le t\le 4$

Calculation:

The displacement of the particle

  $\begin{array}{l}\frac{ds}{dt}=v\left(t\right)\\ s={\displaystyle \int v\left(t\right)}dt\\ s={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{4-t}}dt\\ s={\left|-\frac{2}{3}{\left(4-t\right)}^{3/2}\right|}_0^4\\ s=-\frac{2}{3}\left(0-{\left(4\right)}^{3/2}\right)=\frac{16}{3}\end{array}$

Hence, the displacement of particle is $\frac{16}{3}$ .

Now , displacement = final position − initial position

  $\begin{array}{l}\Delta s=s_f-s_0\\ \frac{16}{3}=s_f-s\left(0\right)\\ \frac{16}{3}=s_f-3\\ s_f=\frac{16}{3}+3=\frac{25}{3}=8.34\end{array}$

Therefore, the final position of the is at $8.34m$ from initial position.

c.

To determine

To find the total distance traveled by the particle.

Answer to Problem 6E

The total distance traveled by the particle is $5.34m$ .

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=\sqrt{4-t},0\le t\le 4$

Calculation:

The total distance traveled by the particle is given by

  $s={\displaystyle \int \left|v\left(t\right)\right|dt}$

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{4}{\int }}\left|\sqrt{4-t}\right|dt}\\ s={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt{4-t}dt}\\ s={\left|-\frac{2}{3}{\left(4-t\right)}^{3/2}\right|}_0^4\\ s=-\frac{2}{3}\left(0-{\left(4\right)}^{3/2}\right)=\frac{16}{3}\end{array}$

therefore , the total distance traveled by the particle is $5.34m$ .