Chapter 8.1, Problem 3E

a.

To determine

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

Answer to Problem 3E

The particle will move right when $0\le t<5$ .

The particle will move left when $5<t\le 10$ .

The particle will stop when $t=5$ .

Explanation of Solution

Given:

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

  $v\left(t\right)=49-9.8t,0\le t\le 10$

Calculation:

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

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

  $\begin{array}{l}v\left(t\right)>0\\ 49-9.8t>0\\ 9.8t-49<0\\ t-\frac{49}{9.8}<0\\ t<\frac{49}{9.8}\\ t<5\\ \because 0\le t\le 10\\ \therefore 0\le t<5\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)<0\\ 49-9.8t<0\\ 9.8t-49>0\\ t-\frac{49}{9.8}>0\\ t>5\\ \because 0\le t\le 10\\ \therefore 5<t\le 10\end{array}$

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

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

  $\begin{array}{l}v\left(t\right)=0\\ 49-9.8t=0\\ 9.8t-49=0\\ t=5\end{array}$

b.

To determine

To find the particle’s displacement for the time interval $0\le t\le 10$ and final position of particle if $s\left(0\right)=3$ .

Answer to Problem 3E

The displacement of particle is $0$ and the final position of the particle is $3m$ 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.

  $v\left(t\right)=49-9.8t,0\le t\le 10$

Calculation:

The displacement of the particle

  $\begin{array}{l}\frac{ds}{dt}=v\left(t\right)\\ s={\displaystyle \int v\left(t\right)}dt\\ s={\displaystyle \underset{0}{\overset{10}{\int }}49-9.8tdt}\\ s=49{\displaystyle \underset{0}{\overset{10}{\int }}dt}-9.8{\displaystyle \underset{0}{\overset{10}{\int }}tdt}\\ s=49{\left|t\right|}_0^{10}-9.8{\left|\frac{t^2}{2}\right|}_0^{10}\\ s=49\left(10-0\right)-4.9\left(100-0\right)\\ s=490-490=0\\ s=0\end{array}$

Hence, the displacement of particle is $0$ .

Now , displacement = final position − initial position

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

Therefore, the final position of the is at $3m$ from initial position.

c.

To determine

To find the total distance traveled by the particle.

Answer to Problem 3E

The total distance traveled by the particle is $245m$ .

Explanation of Solution

Given:

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

  $v\left(t\right)=49-9.8t,0\le t\le 10$

Calculation:

The total distance traveled by the particle is given by

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

Therefore,

  $\begin{array}{l}s={\displaystyle \underset{0}{\overset{10}{\int }}\left|49-9.8t\right|dt}\\ s={\displaystyle \underset{0}{\overset{5}{\int }}49-9.8t}dt+{\displaystyle \underset{5}{\overset{10}{\int }}-\left(49-9.8t\right)}dt\\ s={\left|49t-4.9t^2\right|}_0^5-{\left|49t-4.9t^2\right|}_5^{10}\\ s=245\end{array}$

therefore , the total distance traveled by the particle is $245m$ .