A jet fighter lands on an aircraft carrier. m = 35,000 kg, the force of deceleration is 43,000 N, and the initial velocity is 27 m/s. What distance does it travel during the first 7.5 seconds of deceleration?

Answer 1

F=ma,
a=F/m,
s=v^2/2•a=m•v^2/2•F=
=35000•27^2/2•43000=296.7 m