Chapter 7.1, Problem 6E

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 6E

The particle is moving to the right for $0\le t<4$ , not moving to the left, and stopped at $t=4$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=\sqrt{4-t}$ and the interval $0\le t\le 4$ .

Calculation:

Graph the given function over the given interval.

  

The graph of $v\left(t\right)$ intersects the $t$ -axis at $t=4$ . So, the particle stops at $t=4$ .

The velocity is positive for $0\le t<4$ . So, the particle is moving to the right for $0\le t<4$ .

Conclusion:

The particle is moving to the right for $0\le t<4$ , not moving to the left, and stopped at $t=4$ .

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 6E

The displacement of the particle is $\frac{16}{3}$ . If $s\left(0\right)=3$ , the final position of the particle is $\frac{25}{3}$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=\sqrt{4-t}$ and the interval $0\le t\le 4$ .

Calculation:

Integrate the function $v\left(t\right)=\sqrt{4-t}$ over the interval $0\le t\le 4$ to find the displacement of the particle.

  $\begin{array}{c}\text{Displacement}={\displaystyle \underset{0}{\overset{4}{\int }}\left(\sqrt{4-t}\right)dt}\\ ={\left[-\frac{2}{3}{\left(4-t\right)}^{\frac{3}{2}}\right]}_0^4\\ =\left[-\frac{2}{3}{\left(4-4\right)}^{\frac{3}{2}}-\left(-\frac{2}{3}{\left(4-0\right)}^{\frac{3}{2}}\right)\right]\\ =\left[-\frac{2}{3}{\left(0\right)}^{\frac{3}{2}}-\left(-\frac{2}{3}{\left(4\right)}^{\frac{3}{2}}\right)\right]\\ =-0+\frac{16}{3}\\ =\frac{16}{3}\end{array}$

The displacement of the particle is $\frac{16}{3}$ .

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

  $\begin{array}{c}\text{Final position}=\text{initial position}+\text{ displacement}\\ =3+\frac{16}{3}\\ =\frac{25}{3}\end{array}$

Conclusion:

The displacement of the particle is $\frac{16}{3}$ . If $s\left(0\right)=3$ , the final position of the particle is $\frac{25}{3}$ .

c.

To determine

To find: The total distance travelled by the particle.

Answer to Problem 6E

The total distance traveled by the particle is $\frac{16}{3}$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=\sqrt{4-t}$ and the interval $0\le t\le 4$ .

Calculation:

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

  $\begin{array}{c}\text{Distance traveled}={\displaystyle \underset{0}{\overset{4}{\int }}\left|\sqrt{4-t}\right|dt}\\ ={\displaystyle \underset{0}{\overset{4}{\int }}\left(\sqrt{4-t}\right)dt}\\ =\frac{16}{3}\end{array}$

Conclusion:

The total distance traveled by the particle is $\frac{16}{3}$ .