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In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then 'v' is equal to:

1. √(a1 a2) t
2. (2a1a2/(a1+a2)) t
3. ((a1+a2)/2) t
4. √(2a1a2) t

Correct answer: √(a1 a2) t

Solution

For equal race distance s: t_A=sqrt(2s/a1), t_B=sqrt(2s/a2), v_A=sqrt(2*a1*s), v_B=sqrt(2*a2*s). Then v=v_A-v_B and t=t_B-t_A give v/t = sqrt(a1*a2), so v = sqrt(a1*a2)*t.

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