Chapter 8.1, Problem 7E

a.

To determine

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

Answer to Problem 7E

The particle will move right when $t\in \left[0,\frac{\pi }{2}\right)\cup \left(\frac{3\pi }{2},2\pi \right]$ .

The particle will move left when $\frac{\pi }{2}<t<\frac{3\pi }{2}$ .

The particle will stop when $t=\frac{\pi }{2},\frac{3\pi }{2}$ .

Explanation of Solution

Given:

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

  $v\left(t\right)=e^{\sin t}\cos t,0\le t\le 2\pi$

Calculation:

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

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

  $\begin{array}{l}v\left(t\right)>0\\ e^{\sin t}\cos t>0\\ e^{\sin t}>0\text{ and }\cos t>0\\ \left[0,2\pi \right]\cap \left\{\left[0,\frac{\pi }{2}\right)\cup \left(\frac{3\pi }{2},2\pi \right]\right\}\\ t\in \left[0,\frac{\pi }{2}\right)\cup \left(\frac{3\pi }{2},2\pi \right]\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)<0\\ e^{\sin t}\cos t<0\\ e^{\sin t}<0\text{ or }\cos t<0\\ \frac{\pi }{2}<t<\frac{3\pi }{2}\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)=0\\ e^{\sin t}\cos t=0\\ t=\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

b.

To determine

To find the particle’s displacement for the time interval $0\le t\le 2\pi$ and final position of particle if $s\left(0\right)=3$ .

Answer to Problem 7E

The displacement of particle is $0$ and the final position of the particle is $3m$ 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.

  $v\left(t\right)=e^{\sin t}\cos t,0\le t\le 2\pi$

Calculation:

The displacement of the particle

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

Hence, the displacement of particle is $0$ .

Now , displacement = final position − initial position

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

Therefore, the final position of the is at $3m$ from initial position.

c.

To determine

To find the total distance traveled by the particle.

Answer to Problem 7E

The total distance traveled by the particle is $4.7m$ .

Explanation of Solution

Given:

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

  $v\left(t\right)=e^{\sin t}\cos t,0\le t\le 2\pi$

Calculation:

The total distance traveled by the particle is given by

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

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|e^{\sin t}\cos t\right|dt}\\ s={\displaystyle \underset{0}{\overset{\pi /2}{\int }}(e^{\sin t}\cos t)dt}+{\displaystyle \underset{\pi /2}{\overset{3\pi /2}{\int }}-(e^{\sin t}\cos t)}dt+{\displaystyle \underset{3\pi /2}{\overset{2\pi }{\int }}(e^{\sin t}\cos t)}dt\\ s={\left|e^{\sin t}\right|}_0^{\pi /2}-{\left|e^{\sin t}\right|}_{\pi /2}^{3\pi /2}+{\left|e^{\sin t}\right|}_{3\pi /2}^{2\pi }\\ s=\left(e-1\right)-\left(\frac{1}{e}-e\right)+\left(1-\frac{1}{e}\right)\\ s=2e-\frac{2}{e}=4.7\end{array}$

therefore , the total distance traveled by the particle is $4.7m$ .