Chapter 8.1, Problem 1E

a.

To determine

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

Answer to Problem 1E

The particle will move right when $0\le t<\frac{\pi }{2}\cup \frac{3\pi }{2}<t\le 2\pi$ .

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.

  $\begin{array}{l}v\left(t\right)=5\cos t,0\le t\le 2\pi \\ \end{array}$

Calculation:

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

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

  $\begin{array}{l}v\left(t\right)>0\\ 5\cos t>0\\ \cos t>0\\ \because 0\le t\le 2\pi \\ \therefore 0\le t<\frac{\pi }{2}\cup \frac{3\pi }{2}<t\le 2\pi \end{array}$

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

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

  $\begin{array}{l}v\left(t\right)<0\\ 5\cos t<0\\ \cos t<0\\ \because 0\le t\le 2\pi \\ \therefore \frac{\pi }{2}<t<\frac{3\pi }{2}\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)=0\\ 5\cos t=0\\ \cos t=0\\ \because 0\le t\le 2\pi \\ \therefore t=\frac{\pi }{2},\frac{3\pi }{2}\\ \end{array}$

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 1E

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.

  $\begin{array}{l}v\left(t\right)=5\cos t,0\le t\le 2\pi \\ \end{array}$

Calculation:

The displacement of the particle

  $\begin{array}{l}\frac{ds}{dt}=v\left(t\right)\\ s={\displaystyle \int v\left(t\right)}dt\\ s={\displaystyle \underset{0}{\overset{2\pi }{\int }}5\cos tdt}\\ s=5{\displaystyle \underset{0}{\overset{2\pi }{\int }}\cos tdt}\\ s=5{\left|\sin t\right|}_0^{2\pi }\\ s=5\left(\sin 2\pi -\sin 0\right)=5\left(0-0\right)=0\\ s=0\end{array}$

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 1E

The total distance traveled by the particle is $20m$ .

Explanation of Solution

Given:

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

  $\begin{array}{l}v\left(t\right)=5\cos t,0\le t\le 2\pi \\ \end{array}$

Calculation:

The total distance traveled by the particle is given by

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

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|5\cos t\right|dt}\\ s={\displaystyle \underset{0}{\overset{\pi /2}{\int }}5\cos t}dt+{\displaystyle \underset{\pi /2}{\overset{3\pi /2}{\int }}-5\cos t}dt+{\displaystyle \underset{3\pi /2}{\overset{2\pi }{\int }}5\cos t}dt\\ s={\left|5\sin t\right|}_0^{\pi /2}-{\left|5\sin t\right|}_{\pi /2}^{3\pi /2}+{\left|5\sin t\right|}_{3\pi /2}^{2\pi }\\ s=5\left(1-0\right)-5\left(-1-1\right)+5\left(0-\left(-1\right)\right)\\ s=20\end{array}$

therefore , the total distance traveled by the particle is $20m$ .