Question #2b83b

1 Answer
Jane
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.

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