Chapter 2, Problem 31PQ

A particle’s velocity is given by ${\stackrel{\to }{v}}_y\left(t\right)=-at\hat{j}$, where a = 0.758 m/s² is a constant.

1. a. Describe the particle’s motion. In particular, is it speeding up, slowing down, or maintaining constant speed?
2. b. Find the particle’s velocity at t = 0, t = 10.0 s, and t = 5.00 min.
3. c. Find the particle’s speed at t = 0, t = 10.0 s, and t = 5.00 min.

To determine

The motion of the particle.

Answer to Problem 31PQ

The particle is speeding up.

Explanation of Solution

When the particle is moving, as time increases, the velocity of the particle increases. As velocity increases with time, the motion of the particle can be described as increasing speed.

Therefore, the particle is speeding up.

To determine

The particle’s velocity at $t=0\text{s}$, $t=10.0\text{s}$ and $t=5\text{min}$.

Answer to Problem 31PQ

The particle’s velocity at $t=0\text{s}$, $t=10.0\text{s}$ and $t=5\text{min}$ are $0$, $\underset{\_}{-7.58\widehat{j}\text{m}/\text{s}}$ and $\underset{\_}{-2.27\times {10}^2\widehat{j}\text{m}/\text{s}}$ respectively.

Explanation of Solution

Write the expression for the value of the velocity of the particle in terms of the time taken.

    ${\overrightarrow{v}}_y\left(t\right)=-at\widehat{j}$ (I)

Here, ${\overrightarrow{v}}_y\left(t\right)$ is the velocity, $a$ is the acceleration and $t$ is the time taken.

Conclusion:

Substitute $0\text{s}$ for $t$ and $0.758\text{m}/{\text{s}}^2$ for $a$ in the equation (I).

    $\begin{array}{c}{\overrightarrow{v}}_y\left(0\text{s}\right)=-\left(0.758\text{m}/{\text{s}}^2\right)\left(0\text{s}\right)\widehat{j}\\ =0\end{array}$

Substitute $10.0\text{s}$ for $t$ and $0.758\text{m}/{\text{s}}^2$ for $a$ in the equation (I).

    $\begin{array}{c}{\overrightarrow{v}}_y\left(10.0\text{s}\right)=-\left(0.758\text{m}/{\text{s}}^2\right)\left(10.0\text{s}\right)\widehat{j}\\ =-7.58\widehat{j}\text{m}/\text{s}\end{array}$

Substitute $5\text{min}$ for $t$ and $0.758\text{m}/{\text{s}}^2$ for $a$ in the equation (I).

    $\begin{array}{c}{\overrightarrow{v}}_y\left(5\text{min}\right)=-\left(0.758\text{m}/{\text{s}}^2\right)\left(5\text{min}\right)\widehat{j}\\ =-\left(0.758\text{m}/{\text{s}}^2\right)\left(5\text{min}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\widehat{j}\\ =-2.27\times {10}^2\widehat{j}\text{m}/\text{s}\end{array}$

Therefore, the particle’s velocity at $t=0\text{s}$, $t=10.0\text{s}$ and $t=5\text{min}$ are $0$, $\underset{\_}{-7.58\widehat{j}\text{m}/\text{s}}$ and $\underset{\_}{-2.27\times {10}^2\widehat{j}\text{m}/\text{s}}$ respectively.

To determine

The particle’s speed at $t=0\text{s}$, $t=10.0\text{s}$ and $t=5\text{min}$.

Answer to Problem 31PQ

The particle’s speed at $t=0\text{s}$, $t=10.0\text{s}$ and $t=5\text{min}$ are $0$, $\underset{\_}{7.58\text{m}/\text{s}}$ and $\underset{\_}{2.27\times {10}^2\text{m}/\text{s}}$ respectively.

Explanation of Solution

Write the expression for the value of the speed of the particle in terms of the time taken.

    $v=at$ (II)

Here, $v$ is the speed of the particle.

Conclusion:

Substitute $0\text{s}$ for $t$ and $0.758\text{m}/{\text{s}}^2$ for $a$ in the equation (II).

    $\begin{array}{c}v=\left(0.758\text{m}/{\text{s}}^2\right)\left(0\text{s}\right)\\ =0\end{array}$

Substitute $10.0\text{s}$ for $t$ and $0.758\text{m}/{\text{s}}^2$ for $a$ in the equation (II).

    $\begin{array}{c}v=\left(0.758\text{m}/{\text{s}}^2\right)\left(10.0\text{s}\right)\\ =7.58\text{m}/\text{s}\end{array}$

Substitute $5\text{min}$ for $t$ and $0.758\text{m}/{\text{s}}^2$ for $a$ in the equation (II).

    $\begin{array}{c}v=\left(0.758\text{m}/{\text{s}}^2\right)\left(5\text{min}\right)\\ =\left(0.758\text{m}/{\text{s}}^2\right)\left(5\text{min}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\\ =2.27\times {10}^2\text{m}/\text{s}\end{array}$

Therefore, the particle’s velocity at $t=0\text{s}$, $t=10.0\text{s}$ and $t=5\text{min}$ are $0$, $\underset{\_}{7.58\text{m}/\text{s}}$ and $\underset{\_}{2.27\times {10}^2\text{m}/\text{s}}$ respectively.