Question

A jet plane lands with a speed of 102 m/s and can accelerate at a maximum rate of −5.80 m/s2 as it comes to rest.(a) From the instant the plane touches the runway, what is the minimum time needed before it can come to rest? s

Answer 1

This is an uniformly accelerated motion, then we can use the equation:

`a=(v_f-v_0)/(t)`

In this case the final velocity is 0, the initial velocity is 102 m/s and the acceleration is -5.8 m/s²; plugging these values and solving for t we have:

`\begin{gathered} -5.8=(0-102)/(t) \\ t=(-102)/(-5.8) \\ t=17.59 \end{gathered}`

Therefore, the time it takes to stop is 17.59 s