The maximum velocity attained by the car is:

\frac{α β}{2 (α + β)}

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\frac{α β}{α + β}

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\frac{2 α β}{α + β} t

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\frac{4 α β}{α + β} t

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Solution

The correct option is B $\frac{α β}{α + β}$ t
Let $t_{1}$ & $t_{2}$ be the time acceleration and deceleration respectively.
Total time elapsed $= t$
acceleration $= α$ $u = 0$
deceleration $= β$ $V = u + a t \Rightarrow V = 0 + {α t}_{1}$
Let $V$ be the maximum velocity
$t_{1} + t_{2} = t$
$α = \frac{V}{t_{1}} β = \frac{V}{t_{2}} \Rightarrow t_{1} = \frac{V}{α}; t_{2} = \frac{V}{β}$
$t = t_{1} + t_{2}$
$t = \frac{V}{α} + \frac{V}{β} \Rightarrow t = V (\frac{α + β}{α β})$
$\Rightarrow V = (\frac{α β}{α + β}) t$

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