Chapter 3.2, Problem 37SSC

To determine

To Calculate: The final velocity attained by the plane.

Answer to Problem 37SSC

  $70.7\text{m/s}$

Explanation of Solution

Given:

The plane starts from rest and travels a distance of 500m while being accelerated uniformly from rest at the rate of $5.0{\text{m/s}}^{\text{2}}.$

Formula used:

Final velocity with constant acceleration can be obtained by:

  $v_{final}=v_{initial}+a\Delta t$

Where, $\Delta t$ is the time taken, a is acceleration and v represents velocity.

The distance traveled by an object with a constant acceleration can be obtained by:

  $d_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a\Delta t^2$

Here, d represents displacement, v represents velocity, t represents time and a is the acceleration. Calculation:

The time elapsed in the considered motion will be:

  $\begin{array}{c}{d}_{final}=d_{initial}+v_{initial}\Delta t+\frac{1}{2}a{\left(\Delta t\right)}^2\\ 500=0+0\left(\Delta t\right)+\frac{1}{2}\left(5\right){\left(\Delta t\right)}^2\\ \Delta t=\sqrt{\frac{500\times 2}{5}}\\ =14.14\text{s}\end{array}$

Then, the final velocity of the plane can be calculated as follows:

  $\begin{array}{c}{v}_{final}=v_{initial}+a\Delta t\\ =0+\left(5\times 14.14\right)\\ =70.7\text{m/s}\end{array}$

Conclusion:

Thus, the final velocity of the plane is $70.7\text{m/s}$ .