Question

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?

• A. a = 4 $m{s}^{-2}$, t = 30 s
• B. a = 3$m{s}^{-2}$, t = 30 s
• C. a = 4 $m{s}^{-2}$, t = 40 s
• D. a = 3 $m{s}^{-2}$, t = 40 s

Answer 1

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