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Q.

A car starts from rest, moves with an acceleration ‘a’ and then decelerates at a constant rate of ‘b’ for some time to come to rest. If the total time taken is ‘t’. The maximum velocity of car is given by

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a

$\frac{1}{2} (\frac{a b}{a + b}) t$

b

$\frac{1}{2} (\frac{a b}{a + b}) t^{2}$

c

$(\frac{a b}{a + b}) t$

d

$\frac{a^{2} t}{a + b}$

answer is A.

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Detailed Solution

Let maximum velocity $= V$

Now $V = 0 + a t,$

Similarly $0 = V - b t$

From the above equations we get

$t_{1} = \frac{V}{a}$ and $t_{2} = \frac{V}{b}$

$t = t_{1} + t_{2}$

$t = \frac{V}{a} + \frac{V}{b}$

$t = V (\frac{a + b}{a b})$

$\Rightarrow V = (\frac{a b}{a + b}) t$

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