Question #2b83b

1 Answer
Jan 25, 2018

To stop the plane in minimum time,maximum decceleration has to be applied constantly.

So,we can use v=uat (all the symbols are bearing their conventional meaning)

Given, v=0,u=113.6 and a=6

So, t=113.66 i.e 18.93s

So,during this,if the plane runs for a distance of s,then we can apply v2=u22as

So, s=1.075Km

But,given, the length of the aircraft carrier is only 0.80Km , that means even decelerating it at maximum (by 6ms2),the plane can't be safely landed,it will crash to go out of the carrier.