Answer:
9 m/s^2
Explanation:
initial velocity, u = 0 m/s
Final velocity, v = 45 m/s
time, t = 5 second
Let a be the acceleration of the plane.
Use first equation of motion
v = u + a t
45 = 0 + a x 5
a = 9 m/s^2
Thus, the acceleration of the plane at the time of take off is 9 m/s^2.
