Chapter 7.1, Problem 8E

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 8E

The particle is moving to the right for $0<t\le 3$ , not moving to the left, and stopped at $t=0$ .

Explanation of Solution

Given information:

The function $v(t)=\frac{t}{1+t^2}$ and the interval $0\le t\le 3$ .

Calculation:

Graph the given function over the given interval.

  

The graph of $v\left(t\right)$ intersects the $t$ -axis at $t=0$ . So, the particle stops at $t=0$ .

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

Conclusion:

The particle is moving to the right for $0<t\le 3$ , not moving to the left, and stopped at $t=0$ .

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 8E

The displacement of the particle is $\frac{1}{2}\ln \left(10\right)$ . If $s\left(0\right)=3$ , the final position of the particle is $3+\frac{1}{2}\ln \left(10\right)$ .

Explanation of Solution

Given information:

The function $v(t)=\frac{t}{1+t^2}$ and the interval $0\le t\le 3$ .

Calculation:

Integrate the function $v(t)=\frac{t}{1+t^2}$ over the interval $0\le t\le 3$ to find the displacement of the particle.

  $\begin{array}{c}\text{Displacement}={\displaystyle \underset{0}{\overset{3}{\int }}\left(\frac{t}{1+t^2}\right)dt}\\ ={\left[\frac{1}{2}\ln \left(1+t^2\right)\right]}_0^3\\ =\left[\frac{1}{2}\ln \left(1+3^2\right)-\frac{1}{2}\ln \left(1+0^2\right)\right]\\ =\frac{1}{2}\ln \left(10\right)\end{array}$

The displacement of the particle is $\frac{1}{2}\ln \left(10\right)$ .

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

  $\begin{array}{c}\text{Final position}=\text{initial position}+\text{ displacement}\\ =3+\frac{1}{2}\ln \left(10\right)\end{array}$

Conclusion:

The displacement of the particle is $\frac{1}{2}\ln \left(10\right)$ . If $s\left(0\right)=3$ , the final position of the particle is $3+\frac{1}{2}\ln \left(10\right)$ .

c.

To determine

To find: The total distance travelled by the particle.

Answer to Problem 8E

The total distance traveled by the particle is $\frac{1}{2}\ln \left(10\right)$ .

Explanation of Solution

Given information:

The function $v(t)=\frac{t}{1+t^2}$ and the interval $0\le t\le 3$ .

Calculation:

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

  $\begin{array}{c}\text{Distance traveled}={\displaystyle \underset{0}{\overset{3}{\int }}\left|\frac{t}{1+t^2}\right|dt}\\ ={\displaystyle \underset{0}{\overset{3}{\int }}\left(\frac{t}{1+t^2}\right)dt}\\ =\frac{1}{2}\ln \left(10\right)\end{array}$

Conclusion:

The total distance traveled by the particle is $\frac{1}{2}\ln \left(10\right)$ .