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 +(6-5)*t + (1-2)t^2

= 2 +t -t^2 = 0

t^2 -t -2 = 0

(t-2)(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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