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# In a car race, car A takes a time t less than car B at the finish and passes the finishing point with speed v more than that of the car B. Assuming that both the cars starts from rest and travel with constant accelerations a1 and a2 respectively. So, the value of v will be :(√a1/a2)t(√a2/a1)t(a1√a2)t(√a1a2)t

A
(a1/a2)t
B
(a2/a1)t
C
(a1a2)t
D
(a1a2)t
Solution
Verified by Toppr

#### Consider that A takes t1 second, then according to the given problem, B will take (t1+t) seconds. Further let v1 be the velocity of B at finishing point, then velocity of A will be (v1+v). Writing equations of motion for A and Bv1+v=a1t1 ....(i)v1=a2(t2+t) ....(ii)From equations (i) and (ii), we getv=(a1−a2)t1−a2t .....(iii)Total distance travelled by both the cars is equalSA=SB⇒12a1t21=12a2(t1+t)2⇒t1=√a2t√a1−√a2Substituting this value of t1 in equation (iii), we getv=(√a1a2)t

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