# Question #2b83b

Jan 25, 2018

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

So,we can use $v = u - a t$ (all the symbols are bearing their conventional meaning)

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

So, $t = \frac{113.6}{6}$ i.e $18.93 s$

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

So, $s = 1.075 K m$

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