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Question

A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the total distance travelled by the car is given by


A

12(αβα+β)t2

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B

12(α+βαβ)t2

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C

12(α2+β2α)t2

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D

12(α2β2β)t2

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Solution

The correct option is A

12(αβα+β)t2


Let t1 be the time during which the car accelerates at a rate α. The velocity at the end of time t1 will be v=u+αt1=0+αt1=αt1(u=0)
The time during which the car decelerates is t2=tt1. For this part, the initial velocity is αt1 and the final velocity is zero and the acceleration is β.
α t1=β(tt1)
t1=βtα+β
The distance travelled in time t1 is
s1=ut1+12αt21=12αt21=12α(βα+β)2t2
Also, the distance travelled in time t2 is given by
02=(αt1)22βs22βs2=α2t12=(αβα+β)2t2s2=12β(αα+β)2t2
s1+s2=12α(βα+β)2t2+12β(αα+β)2t2=12αβα+βt2
Hence, the correct choice is (a).


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