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=u-at (all the symbols are bearing their conventional meaning)

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

So, t=113.6/6 i.e 18.93 s

So,during this,if the plane runs for a distance of s,then we can apply v^2=u^2-2as

So, s = 1.075 Km

But,given, the length of the aircraft carrier is only 0.80 Km , that means even decelerating it at maximum (by 6 m/s^2),the plane can't be safely landed,it will crash to go out of the carrier.