Question

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.

Answer 1

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