Chapter 8.1, Problem 4E

a.

To determine

To determine the time when the particle is moving to the right , to the left, and stopped.

Answer to Problem 4E

The particle will move right when $0\le t<1$ .

The particle will move left when $1\le t<2$ .

The particle will stop when $t=1s,2s$ .

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=6t^2-18t+12,0\le t\le 2$

Calculation:

For the particle to move right, the velocity of particle must be greater than $0$ .

Therefore $v\left(t\right)>0$

  $\begin{array}{l}v\left(t\right)>0\\ 6t^2-18t+12>0\\ t^2-3t+2>0\\ \left(t-1\right)\left(t-2\right)>0\\ \because 0\le t\le 2\\ \therefore 0\le t<1\end{array}$

For the particle to move left, the velocity of particle must be less than $0$ ,

Therefore $v\left(t\right)<0$

  $\begin{array}{l}v\left(t\right)<0\\ 6t^2-18t+12<0\\ t^2-3t+2<0\\ \left(t-1\right)\left(t-2\right)<0\\ \because 0\le t\le 2\\ \therefore 1\le t<2\end{array}$

For the particle to stop , its velocity must be equal to $0$

Therefore, $v\left(t\right)=0$

  $\begin{array}{l}v\left(t\right)=0\\ 6t^2-18t+12=0\\ t^2-3t+2=0\\ \left(t-1\right)\left(t-2\right)=0\\ t=1s,2s\end{array}$

b.

To determine

To find the particle’s displacement for the time interval $0\le t\le 2$ and final position of particle if $s\left(0\right)=3$ .

Answer to Problem 4E

The displacement of particle is $4m$ and the final position of the particle is $7m$ from initial position.

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=6t^2-18t+12,0\le t\le 2$

Calculation:

The displacement of the particle

  $\begin{array}{l}\frac{ds}{dt}=v\left(t\right)\\ s={\displaystyle \int v\left(t\right)}dt\\ s={\displaystyle \underset{0}{\overset{2}{\int }}6t^2-18t+12dt}\\ s={\left|6\left(\frac{t^3}{3}\right)-18\left(\frac{t^2}{2}\right)+12t\right|}_0^2\\ s=\left(6\left(\frac{2^3}{3}\right)-18\left(\frac{2^2}{2}\right)+12\times 2\right)-\left(6\left(\frac{0^3}{3}\right)-18\left(\frac{0^2}{2}\right)+12\times 0\right)\\ s=4\end{array}$

Hence, the displacement of particle is $4m$ .

Now , displacement = final position − initial position

  $\begin{array}{l}\Delta s=s_f-s_0\\ 4=s_f-s\left(0\right)\\ 4=s_f-3\\ s_f=7\end{array}$

Therefore, the final position of the is at $7m$ from initial position.

c.

To determine

To find the total distance traveled by the particle.

Answer to Problem 4E

The total distance traveled by the particle is $6m$ .

Explanation of Solution

Given:

The function $v\left(t\right)$ is the velocity in m/sec of a particle moving along x-axis.

  $v\left(t\right)=6t^2-18t+12,0\le t\le 2$

Calculation:

The total distance traveled by the particle is given by

  $s={\displaystyle \int \left|v\left(t\right)\right|dt}$

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{2}{\int }}\left|6t^2-18t+12\right|dt}\\ s={\displaystyle \underset{0}{\overset{1}{\int }}(6t^2-18t+12)}dt+{\displaystyle \underset{1}{\overset{2}{\int }}-(6t^2-18t+12)}dt\\ s={\left|6\left(\frac{t^3}{3}\right)-18\left(\frac{t^2}{2}\right)+12t\right|}_0^1-{\left|6\left(\frac{t^3}{3}\right)-18\left(\frac{t^2}{2}\right)+12t\right|}_1^2\\ s=\left(6\left(\frac{1^3}{3}\right)-18\left(\frac{1^2}{2}\right)+12\times 1\right)-\left(6\left(\frac{0^3}{3}\right)-18\left(\frac{0^2}{2}\right)+12\times 0\right)\\ -\left(6\left(\frac{2^3}{3}\right)-18\left(\frac{2^2}{2}\right)+12\times 2\right)+\left(6\left(\frac{1^3}{3}\right)-18\left(\frac{1^2}{2}\right)+12\times 1\right)\\ s=6\end{array}$

therefore , the total distance traveled by the particle is $6m$ .