a jumbo jet is standing on a runway waiting for takeoff. It must have a velociyty of 45m/s to acheive takeoff. What is the planes acceleration if it takes five seconds for it to reach its takeoff velocity?

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.