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&a2 respectively. Then V is equal to

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a1a2t

2a1a2a1+a2t

a1+a22t

2a1a2

answer is A.

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Detailed Solution

for cars A,B let time taken by A =t0 by the question, vA-vB=v=(a1-a2)t0-a2t---(1) xB=xA=12a1t02=12a2(t0+t)2 square root, a1t0=a2(t0+t) (a1-a2)t0=a2t t0=a2t(a1-a2)---(2) substitute eqn(2) in (1) v=(a1-a2)a2t(a1-a2)-a2t rationalize to get v=a1a2t

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