Chapter 7.1, Problem 4E

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 4E

The particle is moving to the right for $0\le t<1$ , moving to the left in $1<t\le 2$ and stopped at $t=1$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=6t^2-18t+12$ and the interval $0\le t\le 2$ .

Calculation:

Graph the given function over the given interval.

  

The graph of $v\left(t\right)$ intersects the $t$ -axis at $t=1$ . So, the particle stops at this point.

The velocity is positive for $0\le t<1$ . So, the particle is moving to the right for $0\le t<1$ .

The particle is moving to the left $1<t\le 2$ as the velocity is negative.

Conclusion:

The particle is moving to the right for $0\le t<1$ , moving to the left in $1<t\le 2$ and stopped at $t=1$ .

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 4E

The displacement of the particle is 4. If $s\left(0\right)=3$ , the final position of the particle is 7.

Explanation of Solution

Given information:

The function $v\left(t\right)=6t^2-18t+12$ and the interval $0\le t\le 2$ .

Calculation:

Integrate the function $v\left(t\right)=6t^2-18t+12$ over the interval $0\le t\le 2$ to find the displacement of the particle.

  $\begin{array}{c}\text{Displacement}={\displaystyle \underset{0}{\overset{2}{\int }}\left(6t^2-18t+12\right)dt}\\ ={\left[2t^3-9t^2+12t\right]}_0^2\\ =\left(2{\left(2\right)}^3-9{\left(2\right)}^2+12\left(2\right)\right)-\left(2{\left(0\right)}^3-9{\left(0\right)}^2+12\left(0\right)\right)\\ =\left(16-36+24\right)-\left(0-0+0\right)\\ =4\end{array}$

The displacement of the particle is 4.

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

  $\begin{array}{c}\text{Final position}=\text{initial position}+\text{ displacement}\\ =3+4\\ =7\end{array}$

Conclusion:

The displacement of the particle is 4. If $s\left(0\right)=3$ , the final position of the particle is 7.

c.

To determine

To find: The total distance travelled by the particle.

Answer to Problem 4E

The total distance traveled by the particle is 6.

Explanation of Solution

Given information:

The function $v\left(t\right)=6t^2-18t+12$ and the interval $0\le t\le 2$ .

Calculation:

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

  $\begin{array}{c}\text{Distance traveled}={\displaystyle \underset{0}{\overset{2}{\int }}\left|6t^2-18t+12\right|dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\left(6t^2-18t+12\right)dt}+{\displaystyle \underset{1}{\overset{2}{\int }}\left(-\left(6t^2-18t+12\right)\right)dt}\\ ={\left[2t^3-9t^2+12t\right]}_0^1-{\left[2t^3-9t^2+12t\right]}_1^2\\ =\left[\left(2{\left(1\right)}^3-9{\left(1\right)}^2+12\left(1\right)\right)-\left(2{\left(0\right)}^3-9{\left(0\right)}^2+12\left(0\right)\right)\right]-\left[\left(2{\left(2\right)}^3-9{\left(2\right)}^2+12\left(2\right)\right)-\left(2{\left(1\right)}^3-9{\left(1\right)}^2+12\left(1\right)\right)\right]\\ =\left[\left(2-9+12\right)-0\right]-\left[\left(16-36+24\right)-\left(2-9+12\right)\right]\\ =\left[5-0\right]-\left[4-5\right]\\ =5-\left(-1\right)\\ =6\end{array}$

Conclusion:

The total distance traveled by the particle is 6.