Chapter 8.1, Problem 8E

a.

To determine

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

Answer to Problem 8E

The particle will move right when $0<t\le 3$ .

The particle will not move left .

The particle will stop when $t=0$ .

Explanation of Solution

Given:

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

  $v\left(t\right)=\frac{t}{1+t^2},0\le t\le 3$

Calculation:

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

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

  $\begin{array}{l}v\left(t\right)>0\\ \frac{t}{1+t^2}>0\\ t>0\\ \because 0\le t\le 3\\ \therefore 0<t\le 3\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)<0\\ \frac{t}{1+t^2}<0\\ \because t>0\text{ for all values of }t\text{ therefore particle will not move left}\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)=0\\ \frac{t}{1+t^2}=0\\ t=0\end{array}$

b.

To determine

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

Answer to Problem 8E

The displacement of particle is $\frac{\ln (10)}{2}m$ and the final position of the particle is $\frac{\ln (10)}{2}+3$ 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.

  $v\left(t\right)=\frac{t}{1+t^2},0\le t\le 3$

Calculation:

The displacement of the particle

  $\begin{array}{l}\frac{ds}{dt}=v\left(t\right)\\ s={\displaystyle \int v\left(t\right)}dt\\ s={\displaystyle \underset{0}{\overset{3}{\int }}\frac{t}{1+t^2}}dt\\ \text{let }1+t^2=u\\ \text{differentiating both side}\\ 2tdt=du\\ s={\displaystyle \underset{0}{\overset{3}{\int }}\frac{1}{2u}du}\\ s={\left|\frac{\ln \left(1+t^2\right)}{2}\right|}_0^3\\ s=\frac{\ln (10)}{2}\end{array}$

Hence, the displacement of particle is $\frac{\ln (10)}{2}$ .

Now , displacement = final position − initial position

  $\begin{array}{l}\Delta s=s_f-s_0\\ \frac{\ln (10)}{2}=s_f-s\left(0\right)\\ \frac{\ln (10)}{2}=s_f-3\\ s_f=\frac{\ln (10)}{2}+3\end{array}$

Therefore, the final position of the is at $\frac{\ln (10)}{2}+3$ from initial position.

c.

To determine

To find the total distance traveled by the particle.

Answer to Problem 8E

The total distance traveled by the particle is $\frac{\ln (10)}{2}m$ .

Explanation of Solution

Given:

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

  $v\left(t\right)=\frac{t}{1+t^2},0\le t\le 3$

Calculation:

The total distance traveled by the particle is given by

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

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{3}{\int }}\left|\frac{t}{1+t^2}\right|dt}\\ s={\displaystyle \underset{0}{\overset{3}{\int }}\frac{t}{1+t^2}dt}\\ \text{let }1+t^2=u\\ \text{differentiating both side}\\ 2tdt=du\\ s={\displaystyle \underset{0}{\overset{3}{\int }}\frac{1}{2u}du}\\ s={\left|\frac{\ln \left(1+t^2\right)}{2}\right|}_0^3\\ s=\frac{\ln (10)}{2}\end{array}$

therefore , the total distance traveled by the particle is $\frac{\ln (10)}{2}m$ .