Chapter 7.1, Problem 5E

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 5E

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

Explanation of Solution

Given information:

The function $v\left(t\right)=5{\sin }^{2}t\cos t$ and the interval $0\le t\le 2\pi$ .

Calculation:

Graph the given function over the given interval.

  

The graph of $v\left(t\right)$ intersects the $t$ -axis at $t=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi$ . So, the particle stops at $t=0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},2\pi$ .

The velocity is positive for $0<t<\frac{\pi }{2}$ and $\frac{3\pi }{2}<t<2\pi$ . So, the particle is moving to the right for $0<t<\frac{\pi }{2}$ and $\frac{3\pi }{2}<t<2\pi$ .

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

Conclusion:

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

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 5E

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

Explanation of Solution

Given information:

The function $v\left(t\right)=5{\sin }^{2}t\cos t$ and the interval $0\le t\le 2\pi$ .

Calculation:

Integrate the function $v\left(t\right)=5{\sin }^{2}t\cos t$ over the interval $0\le t\le 2\pi$ to find the displacement of the particle.

  $\begin{array}{c}\text{Displacement}={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left(5{\sin }^{2}t\cos t\right)dt}\\ =5{\left[\frac{{\sin }^3\left(t\right)}{3}\right]}_0^{2\pi }\\ =5\left[\frac{{\sin }^3\left(2\pi \right)}{3}-\frac{{\sin }^3\left(0\right)}{3}\right]\\ =5\left[\frac{0}{3}-\frac{0}{3}\right]\\ =0\end{array}$

The displacement of the particle is 0.

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

  $\begin{array}{c}\text{Final position}=\text{initial position}+\text{ displacement}\\ =3+0\\ =3\end{array}$

Conclusion:

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

c.

To determine

To find: The total distance travelled by the particle.

Answer to Problem 5E

The total distance traveled by the particle is $\frac{4}{3}$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=5{\sin }^{2}t\cos t$ and the interval $0\le t\le 2\pi$ .

Calculation:

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

  $\text{Distance traveled}={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|5{\sin }^{2}t\cos t\right|dt}$

  $={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(5{\sin }^{2}t\cos t\right)dt}+{\displaystyle \underset{\frac{\pi }{2}}{\overset{\pi }{\int }}\left(-5{\sin }^{2}t\cos t\right)dt}+{\displaystyle \underset{\pi }{\overset{\frac{3\pi }{2}}{\int }}\left(-5{\sin }^{2}t\cos t\right)dt}+{\displaystyle \underset{\frac{3\pi }{2}}{\overset{2\pi }{\int }}\left(5{\sin }^{2}t\cos t\right)dt}$

  $={\left[\frac{{\sin }^3\left(t\right)}{3}\right]}_0^{\frac{\pi }{2}}-{\left[\frac{{\sin }^3\left(t\right)}{3}\right]}_{\frac{\pi }{2}}^{\pi }-{\left[\frac{{\sin }^3\left(t\right)}{3}\right]}_{\pi }^{\frac{3\pi }{2}}+{\left[\frac{{\sin }^3\left(t\right)}{3}\right]}_{\frac{3\pi }{2}}^{2\pi }$

  $=\left[\frac{{\sin }^3\left(\frac{\pi }{2}\right)}{3}-\frac{{\sin }^3\left(0\right)}{3}\right]-\left[\frac{{\sin }^3\left(\pi \right)}{3}-\frac{{\sin }^3\left(\frac{\pi }{2}\right)}{3}\right]-\left[\frac{{\sin }^3\left(\frac{3\pi }{2}\right)}{3}-\frac{{\sin }^3\left(\pi \right)}{3}\right]+\left[\frac{{\sin }^3\left(2\pi \right)}{3}-\frac{{\sin }^3\left(\frac{3\pi }{2}\right)}{3}\right]$

  $\begin{array}{c}=\left(\frac{1}{3}-\frac{0}{3}\right)-\left(\frac{0}{3}-\frac{1}{3}\right)-\left(\frac{-1}{3}-\frac{0}{3}\right)+\left(\frac{0}{3}-\frac{-1}{3}\right)\\ =\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\\ =\frac{4}{3}\end{array}$

Conclusion:

The total distance traveled by the particle is $\frac{4}{3}$ .