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