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Different Kinds of Motion

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Question

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 accelerated for the same time $t_{1}$ with constant acceleration $2 a_{1}$ and comes 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:

v_{2} = 2 v_{1}

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

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s_{2} = 2 s_{1}

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

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Solution

The correct option is D $2 s_{1} < s_{2} < 4 s_{1}$
$v - t$ graph for case $(1)$ can be considered as,

$V_{m a x} = a_{1} t_{1} = a_{2} t_{2}$
$s_{1} = Area under v-t graph = \frac{1}{2} (t_{1} + t_{2}) a t_{1}$
$V_{a v g} = v_{1} = \frac{s_{1}}{(t_{1} + t_{2})} = \frac{a t_{1}}{2}$

$v - t$ graph for case $(2)$ can be considered as,

$s_{2} = Area under v-t graph = \frac{1}{2} (t_{1} + t_{3}) 2 a t_{1}$
$V_{a v g} = v_{2} = \frac{s_{2}}{(t_{1} + t_{3})} = a t_{1}$

$v_{2} = 2 v_{1}$
Option A is correct.

$s_{2} = (t_{1} + 2 t_{2}) a t_{1} > 2 s_{1} = (t_{1} + t_{2}) a t_{1}$
$s_{2} = (t_{1} + 2 t_{2}) a t_{1} < 4 s_{1} = 2 (t_{1} + t_{2}) a t_{1}$
So, $2 s_{1} < s_{2} < 4 s_{1}$
Option D is correct

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