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Starting from rest a particle is first accelerated for time $t_{1}$ with constant acceleration $a_{1}$ and then stops in time $t_{2}$ with constant retardation $a_{2}$ . Let $v_{1}$ be the average velocity in this case and $s_{1}$ the total displacement. In the second case, it is accelerating for the same time $t_{1}$ with constant acceleration $2 a_{1}$ and come to rest with constant retardation $a_{2}$ in time $t_{3}$ . If $v_{2}$ is the average velocity in this case and $s_{2}$ the total displacement, then

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$2 s_{1} < s_{2} < 4 s_{1}$

$s_{2} = 2 s_{1}$

$v_{2} = 2 v_{1}$

$2 v_{1} < v_{2} < 4 v_{1}$

answer is A, D.

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Detailed Solution

$a_{1} t_{1} = a_{2} t_{2} v_{m a x} = a_{1} t_{1} = a_{2} t_{2}$

$S_{1} =$ Area of $v - t$ graph

$= \frac{1}{2} (t_{1} + t_{2}) (a_{1} t_{1})$

$v_{1} = \frac{S_{1}}{t_{1} + t_{2}} = \frac{1}{2} a_{1} t_{1} S_{2} = \frac{1}{2} (t_{1} + t_{3}) v_{m a x} = \frac{1}{2} (t_{1} + 2 t_{2}) (2 a_{1} t_{1}) v_{2} = \frac{S_{2}}{(t_{1} + t_{3})} = a_{1} t_{1}$

From the four relations, we can see that

$v_{2} = 2 v_{1}$

and $2 S_{1} < S_{2} < 4 S_{1}$

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