a plane has a take off speed of 88.3 m/s and requires 1365 m to reach that speed. determine the acceleration of the plane and the time taken to reach the take off speed.

Question

anonymous · Answer

here, taking initial velocity u=0m/s, final velocity v=88.3m/s and total length s=1365m, so assuming uniform acceleration we apply Newton's laws of motion
$$v ^{2}=u ^{2}+2as$$
$$v ^{2}/2s=a$$ (since u=0m/s) 
So
$$a=2.86ms ^{-2}$$
 Since v=u +at
v/a=t (since u=0m/s)
so t=30.9s