Chapter 7.1, Problem 1E

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 1E

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

Explanation of Solution

Given information:

The function $v\left(t\right)=5\cos t$ 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 $t=\frac{\pi }{2},\frac{3\pi }{2}$ . So, the particle stops at these points.

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

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

Conclusion:

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

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 1E

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 $v\left(t\right)=5\cos t$ and the interval $0\le t\le 2\pi$ .

Calculation:

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

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

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 1E

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

Explanation of Solution

Given information:

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

Calculation:

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

  $\begin{array}{c}\text{Distance traveled}={\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|5\cos t\right|dt}\\ ={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\left(5\cos t\right)dt}+{\displaystyle \underset{\pi /2}{\overset{3\pi /2}{\int }}\left(-5\cos t\right)dt}+{\displaystyle \underset{\pi /2}{\overset{2\pi }{\int }}\left(5\cos t\right)dt}\\ ={\left[5\sin t\right]}_0^{\pi /2}-{\left[5\sin t\right]}_{\pi /2}^{3\pi /2}+{\left[5\sin t\right]}_{3\pi /2}^{2\pi }\\ =\left(5\cdot 1-5\cdot 0\right)-\left(5\cdot \left(-1\right)-5\left(1\right)\right)+\left(5\left(0\right)-5\left(-1\right)\right)\\ =5-\left(-10\right)+5\\ =20\end{array}$

Conclusion:

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