A small airplane starts from rest at one end of a runway and accelerates at 4.0 m/s^2 for 15 s before takeoff.

(a) What is the speed at takeoff?  The answer was provided by the instructor as 60m/s.


I am using the acceleration equation (acceleration = change in velocity/time interval).  The final velocity (4.0 m/s^2), starting velocity (rest which implies is 0), and time interval (15 s) are given in the answer.

1.  Solve the change in velocity by subtracting the 4.0 m/s^2 - 0.  The answer is 4.0 m/s^2.  

2.  Solve the acceleration equation (Acceleration = 4.0 m/s^2  / 15s) 


so how does 4.0 m/s^2 / 15s = 60 m/s?

Guest May 26, 2017

1+0 Answers


Your own formula:


(acceleration = change in velocity/time interval)

4m/s^2 =change in velocity/15s, so

Velocity =4m/s^2 x 15s =60 m/s

Guest May 26, 2017

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