Chapter 7.1, Problem 2E

a.

To determine

To find: When the particle is moving to the right, when moving to the left and when it is stopped.

Answer to Problem 2E

The particle is moving to the right for $0\le t<\frac{\pi }{3}$ , moving to the left in $\frac{\pi }{3}<t\le \frac{\pi }{2}$ and stopped at $t=\frac{\pi }{3}$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=6\sin \left(3t\right)$ and the interval $0\le t\le \frac{\pi }{2}$ .

Calculation:

Graph the given function over the given interval.

  

The graph of $v\left(t\right)$ intersects the $t$ -axis at $t=\frac{\pi }{3}$ . So, the particle stops at this point.

The velocity is positive for $0\le t<\frac{\pi }{3}$ . So, the particle is moving to the right for $0\le t<\frac{\pi }{3}$ .

The particle is moving to the left $\frac{\pi }{3}<t\le \frac{\pi }{2}$ as the velocity is negative.

Conclusion:

The particle is moving to the right for $0\le t<\frac{\pi }{3}$ , moving to the left in $\frac{\pi }{3}<t\le \frac{\pi }{2}$ and stopped at $t=\frac{\pi }{3}$ .

b.

To determine

To find: The particle’s displacement over the given interval, and the final position of the particle if $s\left(0\right)=3$ .

Answer to Problem 2E

The displacement of the particle is $18$ . If $s\left(0\right)=3$ , the final position of the particle is $21$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=6\sin \left(3t\right)$ and the interval $0\le t\le \frac{\pi }{2}$ .

Calculation:

Integrate the function $v\left(t\right)=6\sin \left(3t\right)$ over the interval $0\le t\le \frac{\pi }{2}$ to find the displacement of the particle.

  $\begin{array}{c}\text{Displacement}={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(6\sin \left(3t\right)\right)dt}\\ ={\left[-6\cdot 3\cos \left(3t\right)\right]}_0^{\frac{\pi }{2}}\\ =-18\cos \left(\frac{3\pi }{2}\right)+18\cos \left(0\right)\\ =-18\cdot 0+18\cdot 1\\ =18\end{array}$

The displacement of the particle is $18$ .

If $s\left(0\right)=3$ , the final position of the particle will be:

  $\begin{array}{c}\text{Final position}=\text{initial position}+\text{ displacement}\\ =3+18\\ =21\end{array}$

Conclusion:

The displacement of the particle is $18$ . If $s\left(0\right)=3$ , the final position of the particle is $21$ .

c.

To determine

To find: The total distance travelled by the particle.

Answer to Problem 2E

The total distance traveled by the particle is $54$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=6\sin \left(3t\right)$ and the interval $0\le t\le \frac{\pi }{2}$ .

Calculation:

Find the total distance traveled by the particle over the given interval as follows:

  $\begin{array}{c}\text{Distance traveled}={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\left|6\sin \left(3t\right)\right|dt}\\ ={\displaystyle \underset{0}{\overset{\pi /3}{\int }}\left(6\sin \left(3t\right)\right)dt}+{\displaystyle \underset{\pi /3}{\overset{\pi /2}{\int }}\left(-6\sin \left(3t\right)\right)dt}\\ ={\left[-18\cos \left(3t\right)\right]}_0^{\pi /3}+{\left[18\cos \left(3t\right)\right]}_{\pi /3}^{\pi /2}\\ =-18\left(\cos \left(\pi \right)-\cos \left(0\right)\right)+\left(18\cos \left(\frac{3\pi }{2}\right)-18\cos \left(\pi \right)\right)\\ =-18\left(\left(-1\right)-1\right)+18\left(0-\left(-1\right)\right)\\ =-18\left(-2\right)+18\left(1\right)\\ =54\end{array}$

Conclusion:

The total distance traveled by the particle is $54$ .