Starting from rest and accelerating at 4.8 m/s², the plane's velocity at time t is
v(t) = (4.8 m/s²) t
so that after 15 s, its speed at takeoff be
v(15 s) = (4.8 m/s²) * (15 s) = 72 m/s
In this time it takes for the plane to take off, it will travel a distance at time t of
x(t) = 1/2 (4.8 m/s²) t²
and it takes 15 s to take off, so the runway would have to have a length of
x(15 s) = 1/2 (4.8 m/s²) * (15 s)² = 540 m
