Chapter 3, Problem 94A

(a)

To determine

The initial acceleration of the sled and compare it with the magnitude of the free-fall acceleration.

Answer to Problem 94A

The acceleration of the sled at its starting is $247{\text{m/s}}^2$ which is $25$ times the magnitude of the free-fall acceleration.

Explanation of Solution

Given:

The initial velocity of the sled is $v_{\text{i}}=0\text{m/s}$ .

The final velocity of the sled is $v_{\text{f}}=444\text{m/s}$ .

The time interval for speed up is $\Delta t=1.8\text{s}$ .

Formula used:

Show the expression for the acceleration $\left(\overline{a}\right)$ of an object as follows:

  $\begin{array}{l}\overline{a}=\frac{\Delta v}{\Delta t}\\ \overline{a}=\frac{v_{\text{f}}-v_{\text{i}}}{\Delta t}\end{array}$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ are the initial and final velocity of the object, and $\Delta t$ is the time interval.

Calculation:

Find the initial acceleration of the sled as follows:

  $\overline{a}=\frac{v_{\text{f}}-v_{\text{i}}}{\Delta t}$

Apply the values for $v_{\text{i}}$ , $v_{\text{f}}$ , and $\Delta t$ in the above expression,

  $\begin{array}{l}\overline{a}=\frac{444-0}{1.8}\\ \overline{a}=246.67{\text{m/s}}^2\\ \overline{a}\approx 247{\text{m/s}}^2\end{array}$

Hence, the acceleration of the sled at the starting is $247{\text{m/s}}^2$ .

The magnitude of the free-fall acceleration $\left(g\right)$ is $9.81{\text{m/s}}^2$ .

Compare the acceleration of the sled with the magnitude of the free-fall acceleration.

  $\begin{array}{l}\frac{\overline{a}}{g}=\frac{247{\text{m/s}}^2}{9.81{\text{m/s}}^2}\\ \frac{\overline{a}}{g}=25\\ \overline{a}=25g\end{array}$

Conclusion:

Thus, the acceleration of the sled at its starting is $247{\text{m/s}}^2$ which is $25$ times the magnitude of the free-fall acceleration.

(b)

To determine

The acceleration of the sled at its braking and compare it with the magnitude of the free-fall acceleration.

Answer to Problem 94A

The acceleration of the sled at its breaking is $247{\text{m/s}}^2$ which is $21$ times the magnitude of the free-fall acceleration.

Explanation of Solution

Given:

A sled moving with a velocity $v_{\text{i}}$ comes to stop in the time interval $\Delta t$ .

The initial velocity of the sled is $v_{\text{i}}=444\text{m/s}$ .

The final velocity of the sled is $v_{\text{f}}=0\text{m/s}$ .

The time interval is $\Delta t=2.15\text{s}$ .

Formula used:

Show the expression for the acceleration $\left(\overline{a}\right)$ of an object as follows:

  $\begin{array}{l}\overline{a}=\frac{\Delta v}{\Delta t}\\ \overline{a}=\frac{v_{\text{f}}-v_{\text{i}}}{\Delta t}\end{array}$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ are the initial and final velocity of the object, and $\Delta t$ is the time interval.

Calculation:

Find the initial acceleration of the sled as follows:

  $\overline{a}=\frac{v_{\text{f}}-v_{\text{i}}}{\Delta t}$

Apply the values for $v_{\text{i}}$ , $v_{\text{f}}$ , and $\Delta t$ in the above expression,

  $\begin{array}{l}\overline{a}=\frac{0-444}{2.15}\\ \overline{a}=-206.51{\text{m/s}}^2\\ \overline{a}\approx -207{\text{m/s}}^2\end{array}$

Hence, the acceleration of the sled at its braking is $-207{\text{m/s}}^2$ .

The negative sign indicates retardation of the sled.

Compare the acceleration of the sled with the magnitude of the free-fall acceleration.

  $\begin{array}{l}\frac{\overline{a}}{g}=\frac{207{\text{m/s}}^2}{9.81{\text{m/s}}^2}\\ \frac{\overline{a}}{g}=21\\ \overline{a}=21g\end{array}$

Conclusion:

Thus, the acceleration of the sled at its breaking is $-207{\text{m/s}}^2$ which is $21$ times the magnitude of the free-fall acceleration.