physics
posted by Marissa on .
At the moment car A is starting from rest and accelerating at 5.0 m/s2, car B passes it moving at a constant speed of 45m/s. How long will it take car A to catch up with car B?

A will pass B when the distances travelled by each are the same, at the same time t.
(a/2) t^2 = V*t
t = 0 or 2 V/a = 18 seconds 
Since know one has answered this question, I will give it a try. But hopefully someone checks my work because I am a little tired. For Car A, Vi=0, a=5m/s^2, and t=?. For Car B, Vi=45m/s, Vf=45m/s, a=0, and t=?. Using two of the kinematic equations, 1/2(Vi+Vf)t=d and d=Vi(t)+1/2at^2 plug in the values and set the equations equal to each other and solve for t. 1/2(Vi+Vf)t=Vi(t)+1/2at^2. Hopefully you get 18s if I did it correctly. For the equation on the left substitute the values for Car B into it and the equation on the right substitute the values for Car A into it.

I know one who answered it.