An airplane travels down a runway for 4.0 seconds with an acceleration of 9.0 m/s what is its change in velocity during this time

Acceleration is given in the units of meters per second per second (m/s^2). Thus if the airplane is experiencing a constant acceleration of 9 m/s^2 for a known time period of 4 seconds, its total increase in velocity can be found by multiplying 9 m/s^2 * 4s to yield a velocity increase over the period of 36 m/s.

Given: Time t = 4.0 seconds

acceleration a = 9.0 m/s²

Initial Velocity Vi = 0

Required: Change in velocity Vf = ? F

Formula: a = Vf - Vi/t

Vf = Vi + at

Vf = 0 m/s + (9.0 m/s²) (4.0 s)

Vf = 36 m/s