We investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on solid ground with a speed of 115 m/s, and its acceleration can have a maximum magnitude of 8.97 m/s2 as it comes to rest. (a) From the instant the jet touches the runway, what is the minimum time interval needed before it can come to rest? (b) Can this jet land on a small tropical island airport where the runway is 0.800 km long?

If we use the maximum acceleration it will give us the results for the minimum time interval possible and the minimum distance needed to come to a stop.

We start with the equation [tex]v=v_0+at[/tex], which means:

[tex]t=\frac{v-v_0}{a}[/tex]

We will get the minimum time interval by using the maximum acceleration (decceleration in this case) that makes the jet go from [tex]v_0=115m/s[/tex] to [tex]v=0m/s[/tex]. Taking the direction of travel as positive (and thus negative acceleration), we get:

[tex]t=\frac{(0m/s)-(115m/s)^2}{-8.97m/s^2}=12.82s[/tex]

The distance traveled with maximum acceleration can be calculated with the formula [tex]v^2=v_0^2+2ad[/tex], which means:

[tex]d=\frac{v^2-v_0^2}{2a}[/tex]

Again taking the direction of travel as positive we get:

[tex]d=\frac{(0m/s)^2-(115m/s)^2}{2(-8.97m/s^2)}=737.18m[/tex]

Which means that for a runway 0.8km=800m long it will be enough.