Question

An aeroplane taking off from a field has a run 500m . What is the acceleration and take off velocity if it leaves the underground 10 s after the start

Answer 1

here, distance s=500
initial velocity u=0
time t=10
now,

$s=\frac{(u+v)t}{2}$
$s=\frac{vt}{2}$
$v=\frac{2s}{t}=2*500/10=100$
so take off velocity is 100 m/s

again,

$v=u+at$
$v=at$
$a=\frac{v}{t}=100/10=10$
so acceleration is 10 m/s^2