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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 $$\beta$$ and comes to rest. It total time elapsed is $$t$$, then maximum velocity acquired by car will be:


A
(α2β2)tαβ
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B
(α2+β2)tαβ
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C
(α+β)tαβ
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D
(αβ)tα+β
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Solution

The correct option is D $$\cfrac { \left( { \alpha }^{ }{ \beta }^{ } \right) t }{ \alpha +\beta } $$
From rest, the car acceleraters for time t,
So $$\alpha = \dfrac{V}{t_1} $$       ...(1) , let maximum velocity is V.
The car begins decelerate after attaining maximum velocity
$$\beta = \dfrac{V}{t-t_1}$$            ....(2)

By adding 1 and 2

$$t = v(\frac{1}{\alpha} + \dfrac{1}{\beta}$$

$$v = \dfrac{(\alpha \beta)t}{\alpha + \beta} $$

Hence (D) is correct answer


Physics

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