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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 of journey is t, then the maximum velocity acquired by the car is given by

A
(α+βαβ)t
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B
αβα+βt
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C
α2β2αβt
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D
αβαβt
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Solution

The correct option is B αβα+βt
Let the car accelerates for time t1 and decelerates for time t2. Then t1+t2=t
For the first journey
Vmax=0+at1
t1=Vmaxα
For the second journey
0=Vmaxβt2
t2=Vmaxβ
Then the total time taken is
Vmaxα+Vmaxβ=t
Vmax=αβtα+β
Therefore option b is correct.

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