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Question

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/s
Given 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=v1v

Velocity of car B is given by

v2=a2t2 (from equation (i))

v1v=a2(t1+20)

v1v=40(t1+20) (ii)

Subtracting equation (ii) from equation (i)

v1(v1v)=50t140(t1+20)

v=10t1800 (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)

(2520)t21800t18000=0

Solving the quadratic equation, we get:

t1=169.44 and 9.44 (Neglected)

Substitute t1 in equation (iii)

v=10t1800

v=10×169.44800

v=894.4 m/s894 m/s

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