Physics
posted by Ajani on .
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?

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. 
Thank you! i have spent an hour on that question. Which equation was that?
s=ut+at^2/2 ? 
Yes, the equation for Xa  Xb results from
s = so + uo*t + a*t^2/2
where so is the displacement at t = 0