Chapter 3, Problem 66A

To determine

The displacement of the plane that has uniform acceleration.

Answer to Problem 66A

  $924.048\text{m}$

Explanation of Solution

Given:

Initial and final velocity of the plane is $66\text{m/s}$ and $88\text{m/s}$ , respectively. And, time taken by the plane is 12s.

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:

Using the given values of velocities and time in above formula, the acceleration of the plane can be calculated as:

  $\begin{array}{l}88=66+a\left(12\right)\\ a=1.834{\text{m/s}}^{\text{2}}\end{array}$

Then, the displacement of the plane will be,

  $\begin{array}{c}{d}_f=0+\left(66\right)\left(12\right)+\frac{1}{2}\left(1.834\right){\left(12\right)}^2\\ =792+132.048\\ =924.048\text{m}\end{array}$

Conclusion:

Thus, the displacement of the plane is $924.048\text{m}$ .