Question

Question #2b83b

Answer 1

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.