Chapter 8.1, Problem 10E

a.

To determine

To determine the displacement of particle .

Answer to Problem 10E

The displacement of particle $-1.449m$

Explanation of Solution

Given:

A particle travels with velocity $v\left(t\right)=\left(t-2\right)\sin tm/\sec \text{ for }0\le t\le 4\text{ sec}$ .

Calculation:

  $\begin{array}{l}v\left(t\right)=\left(t-2\right)\sin t\\ s={\displaystyle \underset{0}{\overset{4}{\int }}\left(t-2\right)\sin tdt}\\ s={\displaystyle \underset{0}{\overset{4}{\int }}t\sin tdt}-{\displaystyle \underset{0}{\overset{4}{\int }}2\sin tdt}\\ s={\left|-t\cos t+\sin t\right|}_0^4-2{\left|-\cos t\right|}_0^4\\ s=-1.449\end{array}$

Hence, displacement of particle $-1.449m$

b.

To determine

To find the total distance travel by a particle.

Answer to Problem 10E

The total distance traveled by particle is $1.914m$ .

Explanation of Solution

Given:

A particle travels with velocity $v\left(t\right)=\left(t-2\right)\sin tm/\sec \text{ for }0\le t\le 4\text{ sec}$ .

Calculation:

  $\begin{array}{l}v\left(t\right)=\left(t-2\right)\sin t\\ s={\displaystyle \underset{0}{\overset{4}{\int }}\left|\left(t-2\right)\sin t\right|dt}\\ \text{To evaluate integral , the critical points at which }\left(t-2\right)\sin t\text{ changes sign must be known}\\ \text{so,}\\ \left(t-2\right)\sin t=0\\ t-2=0\text{ and }\sin t=0\\ t=2\text{ , }0,\pi \\ \because 0\le t\le 4\\ \text{therefore, }\\ s={\displaystyle \underset{0}{\overset{2}{\int }}\left(t-2\right)\sin tdt}+{\displaystyle \underset{2}{\overset{\pi }{\int }}\left(t-2\right)\sin tdt}+{\displaystyle \underset{\pi }{\overset{4}{\int }}\left(t-2\right)\sin tdt}\\ s={\left|-t\cos t+2\cos t-2\sin t\right|}_0^2+{\left|-t\cos t+2\cos t-2\sin t\right|}_2^{\pi }+{\left|-t\cos t+2\cos t-2\sin t\right|}_{\pi }^4\\ s=1.914\end{array}$

Hence , the total distance traveled by particle is $1.914m$ .