Chapter 3, Problem 91A

(a)

To determine

The acceleration provided by the gun to the object.

Answer to Problem 91A

  $\overline{a}=3.1\times {10}^8{\text{m/s}}^2$

Explanation of Solution

Given:

Consider a small object is released from the gun.

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

The final velocity of the object is $v_{\text{f}}=3.5\text{km/s}$ .

The distance traveled by the object after releasing from the gun is $d=2\text{cm}$ .

Consider the acceleration provided by the gun to the object is $\overline{a}$ .

Formula used:

Equation of motion for uniform acceleration: Consider an object is moving with uniform acceleration, then the equation of motion can be given as follows:

  $\begin{array}{l}{v}_{\text{f}}^2-v_{\text{i}}^2=2\overline{a}\left(d_{\text{f}}-d_{\text{i}}\right)\\ v_{\text{f}}^2-v_{\text{i}}^2=2\overline{a}d\end{array}$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ are the initial and final velocity of the object, $d_{\text{i}}$ and $d_{\text{f}}$ are the initial and final distance covered by the object, $\overline{a}$ is the acceleration of the object, and $d$ is the distance traveled by the object.

Calculation:

Find the acceleration provided by the gun to the object as follows:

Apply the equation of motion,

  $\begin{array}{c}{v}_{\text{f}}^2-v_{\text{i}}^2=2\overline{a}d\\ {\left(3.5\text{km/s}\times \frac{1000\text{m/s}}{1\text{km/s}}\right)}^2-0^2=2\overline{a}\times \left(2\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ {\left(3500\text{m/s}\right)}^2=2\overline{a}\times \left(0.02\text{m}\right)\\ 12,250,000=0.04\overline{a}\\ \overline{a}=\left(\frac{12,250,000}{0.04}\right)\\ \overline{a}=3.1\times {10}^8{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the acceleration provided by the gun to the object is $\overline{a}=3.1\times {10}^8{\text{m/s}}^2$ .

(b)

To determine

The time interval for the acceleration to take place.

Answer to Problem 91A

  $11\mu \text{s}$

Explanation of Solution

Given info:

Consider a small object is released from the gun.

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

The final velocity of the object is $v_{\text{f}}=3.5\text{km/s}$ .

The distance traveled by the object after releasing from the gun is $d=2\text{cm}$ .

Consider the time taken interval for the acceleration to take place is $t$ .

Consider the acceleration provided by the gun to the object is $\overline{a}$ .

Formula used:

Equation of Motion for uniform acceleration:Consider an object is moving with uniform acceleration, then the equation of motion can be given as follows:

  $v_{\text{f}}=v_{\text{i}}+\overline{a}t$

Here, $v_{\text{i}}$ and $v_{\text{f}}$ are the initial and final velocity of the object, $t$ is the time interval, $\overline{a}$ is the acceleration of the object, and $d$ is the distance traveled by the object.

Calculation:

Find the time taken for the acceleration to take place as follows:

Apply Equation of Motion,

  $\begin{array}{c}{v}_{\text{f}}=v_{\text{i}}+\overline{a}t\\ \left(3.5\text{km/s}\times \frac{1000\text{m/s}}{1\text{km/s}}\right)=0+\left(3.1\times {10}^8{\text{m/s}}^2\right)\times t\\ 3500=3.1\times {10}^{8}t\\ t=\left(\frac{3500}{3.1\times {10}^8}\right)\\ t=11\times {10}^{-6}\text{s}\\ t=11\mu \text{s}\end{array}$

Conclusion:

Thus, the time taken for the acceleration to take place is $11\mu \text{s}$ .