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# In a car race, car A takes 20 s less than car B to finish. It passes the finishing point with speed v more than that of car B. Assuming that both the cars start from rest and travel with a constant acceleration of 50 m/s2 and 40 m/s2 respectively, what is the value of v ?

A
620 m/s
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B
894 m/s
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C
682 m/s
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D
864 m/s
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Solution

## The correct option is B 894 m/sGiven that, Acceleration of car A, a1=50 m/s2 Acceleration of car B, a2=40 m/s2 Initial velocity of both cars, u=0 m/s Let car A take t1 time to travel from rest to destination, reaching with velocity v1. We know that v=u+at ⇒v1=50 m/s2×t1 ……(i) Time taken by car B, t2=t1+20 sec Velocity of car B, v2=v1−v Velocity of car B is given by v2=a2t2 (from equation (i)) ⇒v1−v=a2(t1+20) ⇒v1−v=40(t1+20) ……(ii) Subtracting equation (ii) from equation (i) ⇒v1−(v1−v)=50t1−40(t1+20) ⇒v=10t1−800 ……(iii) Now, total distance travelled by both the cars is the same. So, S1=S2 From S=ut+12at2 12×50×t21=12×40×(t1+20)2 ⇒25×t21=20×(t1+20)2 ⇒25t21=20×(t21+40t1+400) ⇒(25−20)t21−800t1−8000=0 Solving the quadratic equation, we get: t1=169.44 and −9.44 (Neglected) Substitute t1 in equation (iii) ⇒v=10t1−800 ⇒v=10×169.44−800 ⇒v=894.4 m/s≅894 m/s  Suggest Corrections  0      Similar questions  Explore more