Posted by **Marissa** on Thursday, January 17, 2013 at 10:42pm.

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?

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**drwls**, Thursday, January 17, 2013 at 11:29pm
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

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**Devron**, Thursday, January 17, 2013 at 11:39pm
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.

- physics -
**drwls**, Friday, January 18, 2013 at 1:21am
I know one who answered it.

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