Posted by Ajani on Thursday, May 17, 2012 at 6:42am.
At an instant, Runner 'A' runs at 6m/s with an acceleration of 2m/s^2. Runner 'B' runs at 5m/s with an acceleration of 4m/s^2, and the former (Runner A) is 2 metres ahead of the latter (runner B). How long does it take for Runner B to catch up with runner A?

Physics  drwls, Thursday, May 17, 2012 at 6:51am
Start time measurement from when the separation distance, Xa  Xb, is 2.0 m.
Xa  Xb = 2.0 +(65)*t + (12)t^2
= 2 +t t^2 = 0
t^2 t 2 = 0
(t2)(t+1) = 0
Solve for t. Take the positive root.
t = 2.0 s.

Physics  Ajani, Thursday, May 17, 2012 at 6:56am
Thank you! i have spent an hour on that question. Which equation was that?
s=ut+at^2/2 ?

Physics  drwls, Thursday, May 17, 2012 at 8:13am
Yes, the equation for Xa  Xb results from
s = so + uo*t + a*t^2/2
where so is the displacement at t = 0
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