Chapter 7.1, Problem 3E

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 3E

The particle is moving to the right for $0\le t<5$ , moving to the left in $5<t\le 10$ and stopped at $t=5$ .

Explanation of Solution

Given information:

The function $v\left(t\right)=49-9.8t$ and the interval $0\le t\le 10$ .

Calculation:

Graph the given function over the given interval.

  

The graph of $v\left(t\right)$ intersects the $t$ -axis at $t=5$ . So, the particle stops at this point.

The velocity is positive for $0\le t<5$ . So, the particle is moving to the right for $0\le t<5$ .

The particle is moving to the left $5<t\le 10$ as the velocity is negative.

Conclusion:

The particle is moving to the right for $0\le t<5$ , moving to the left in $5<t\le 10$ and stopped at $t=5$ .

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 3E

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

Explanation of Solution

Given information:

The function $v\left(t\right)=49-9.8t$ and the interval $0\le t\le 10$ .

Calculation:

Integrate the function $v\left(t\right)=49-9.8t$ over the interval $0\le t\le 10$ to find the displacement of the particle.

  $\begin{array}{c}\text{Displacement}={\displaystyle \underset{0}{\overset{10}{\int }}\left(49-9.8t\right)dt}\\ ={\left[49t-4.9t^2\right]}_0^{10}\\ =\left(49\left(10\right)-4.9{\left(10\right)}^2\right)-\left(49\left(0\right)-4.9{\left(0\right)}^2\right)\\ =490-490-0\\ =0\end{array}$

The displacement of the particle is 0.

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

  $\begin{array}{c}\text{Final position}=\text{initial position}+\text{ displacement}\\ =3+0\\ =3\end{array}$

Conclusion:

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

c.

To determine

To find: The total distance travelled by the particle.

Answer to Problem 3E

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

Explanation of Solution

Given information:

The function $v\left(t\right)=49-9.8t$ and the interval $0\le t\le 10$ .

Calculation:

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

  $\begin{array}{c}\text{Distance traveled}={\displaystyle \underset{0}{\overset{10}{\int }}\left|49-9.8t\right|dt}\\ ={\displaystyle \underset{0}{\overset{5}{\int }}\left(49-9.8t\right)dt}+{\displaystyle \underset{5}{\overset{10}{\int }}\left(-\left(49-9.8t\right)\right)dt}\\ ={\left[49t-4.9t^2\right]}_0^5-{\left[49t-4.9t^2\right]}_5^{10}\\ =\left[\left(49\left(5\right)-4.9{\left(5\right)}^2\right)-\left(49\left(0\right)-4.9{\left(0\right)}^2\right)\right]-\left[\left(49\left(10\right)-4.9{\left(10\right)}^2\right)-\left(49\left(5\right)-4.9{\left(5\right)}^2\right)\right]\\ =\left[\left(245-122.5\right)-0\right]-\left[\left(490-490\right)-\left(245-122.5\right)\right]\\ =122.5-\left(-122.5\right)\\ =245\end{array}$

Conclusion:

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