Chapter 8.1, Problem 2E

a.

To determine

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

Answer to Problem 2E

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

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

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

Explanation of Solution

Given:

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

  $v\left(t\right)=6\sin 3t,0\le t\le \pi /2$

Calculation:

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

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

  $\begin{array}{l}v\left(t\right)>0\\ 6\sin 3t>0\\ \sin 3t>0\\ \because 0\le t\le \pi /2\\ \therefore 0\le t<\pi /3\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)<0\\ 6\sin 3t<0\\ \sin 3t<0\\ \because 0\le t\le \pi /2\\ \therefore \frac{\pi }{3}<t\le \frac{\pi }{2}\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)=0\\ 6\sin 3t=0\\ \sin 3t=0\\ \because 0\le t\le \pi /2\\ \therefore t=\frac{\pi }{3}\\ \end{array}$

b.

To determine

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

Answer to Problem 2E

The displacement of particle is $2m$ and the final position of the particle is $5m$ 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.

  $v\left(t\right)=6\sin 3t,0\le t\le \pi /2$

Calculation:

The displacement of the particle

  $\begin{array}{l}\frac{ds}{dt}=v\left(t\right)\\ s={\displaystyle \int v\left(t\right)}dt\\ s={\displaystyle \underset{0}{\overset{\pi /2}{\int }}6\sin 3tdt}\\ s=6{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\sin 3tdt}\\ s=6{\left|-\frac{\cos 3t}{3}\right|}_0^{\pi /2}=-2{\left|\sin 3t\right|}_0^{\pi /2}=-2\left(\sin \frac{3\pi }{2}-\sin 0\right)=2\\ s=2\end{array}$

Hence, the displacement of particle is $2m$ .

Now , displacement = final position − initial position

  $\begin{array}{l}\Delta s=s_f-s_0\\ 2=s_f-s\left(0\right)\\ 2=s_f-3\\ s_f=5\end{array}$

Therefore, the final position of the is at $5m$ from initial position.

c.

To determine

To find the total distance traveled by the particle.

Answer to Problem 2E

The total distance traveled by the particle is $6m$ .

Explanation of Solution

Given:

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

  $v\left(t\right)=6\sin 3t,0\le t\le \pi /2$

Calculation:

The total distance traveled by the particle is given by

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

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\left|6\sin 3t\right|dt}\\ s={\displaystyle \underset{0}{\overset{\pi /3}{\int }}6\sin 3t}dt+{\displaystyle \underset{\pi /3}{\overset{\pi /2}{\int }}-6\sin 3t}dt\\ s={\left|-2\cos 3t\right|}_0^{\pi /3}-{\left|-2\cos 3t\right|}_{\pi /3}^{\pi /2}\\ s=-2\left(\cos \pi -\cos 0\right)+2\left(\cos 3\pi /2-\cos \pi \right)\\ s=6\end{array}$

therefore , the total distance traveled by the particle is $6m$ .