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Q.

A particle starts from rest with constant acceleration $a$ and it is then decelerated with a constant value b till it is brought to rest. The total time taken between these two rest positions is t. What is the maximum speed acquired by the particle?

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a

$(a - b) \frac{t}{2}$

b

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

c

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

d

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

answer is C.

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Detailed Solution

Question Image
$\begin{array}{l} \frac{v}{t_{1}} = a; \frac{v}{t_{2}} = b \\ t_{1} + t_{2} = t \\ v = (\frac{a b}{a + b}) t \end{array}$

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