Question

A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the jet is moving at a speed of 400 m/s. What is the jet’s acceleration, assuming it to be a constant acceleration?

Short answer

The acceleration of the plane is$8.75\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.

Step-by-step solution

Given information

The initial speed of a plane is $v_{is}=300\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Distance covered by plane is $s=4km=4000m$

The final speed of the plane is$v_{fs}=400\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The acceleration of the plane.

Use kinematical equation ${v^2}_{fs}={v^2}_{is}+2as$.

Substitute known values

$$\begin{gathered} {400}^2={300}^2+2a\left(4000\right) \\ {400}^2-{300}^2=8000a \\ a=\frac{70000}{8000} \\ =8.75\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Therefore, the acceleration of the plane is$8.75\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$.