A plane has a take-off speed of 120 $m{s}^{-1}$ and it reaches that speed after covering 2400 m. Determine the acceleration of the plane (assuming it to be constant) and the time required to reach this speed?

The correct option is D

a = 3 $m{s}^{-2}$, t = 40 s

Initial velocity, u = 0

Final velocity, v = 120 $m{s}^{-1}$ and distance s = 2400 m

By the third equation of motion, $v^2=u^2+2as$

${120}^2$= 0 + $2\times a\times 2400$

Solving we get, a = 3 $m{s}^{-2}$

Now by first equation of motion,
v = u + at

120 = 0 + 3 t

t = 40 s