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Question

A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ to come to rest. If the total time elapsed is $$t$$, the maximum velocity acquired by the car is given by


A
(α2+β2αβ)t
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B
(α2β2αβ)t
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C
(α+βαβ)t
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D
(αβα+β)t
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Solution

The correct option is D $$\left( \cfrac { \alpha \beta } { \alpha +\beta } \right)t $$
Let maximum velocity=$$v$$ 
Now,$$ v=0+\alpha { t }_{ 1 }$$
Similarly,$$\ 0=v-\beta { t }_{ 2 }$$
From the above equations we get, $$\\ { t }_{ 1 }=\dfrac { v }{ \alpha  } \quad \& \quad { t }_{ 2 }=\dfrac { v }{ \beta  } \\ { t }_{ 1 }+{ t }_{ 2 }=t=\dfrac { v }{ \alpha  } +\dfrac { v }{ \beta  } \\ \Rightarrow v=\dfrac { \alpha \beta  }{ \alpha +\beta  } t$$

Physics

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