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1st Equation 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 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

A

v_{2} = 2 v_{1}

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B

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

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C

S_{2} 2 = S_{1}

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D

2 S_{1} < S_{2} < 4 S_{1}

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Solution

The correct options are
A $v_{2} = 2 v_{1}$
D $2 S_{1} < S_{2} < 4 S_{1}$
Velocity after acceleration by $a$ for $t_{1}$ time= $a_{1} t_{1}$
It takes time $t_{2}$ to retard to zero velocity, so, $a_{1} t_{1} = a_{2} t_{2}$
$⟹ t_{2} = \frac{a_{1}}{a_{2}} t_{1}$
Similarly for second case,
$t_{3} = \frac{2 a_{1}}{a_{2}} t_{1}$
In first case,
Distance traveled during acceleration= $\frac{1}{2} a_{1} t_{1}^{2}$
Distance traveled during retardation= $\frac{(a_{1} t_{1})^{2}}{2 a_{2}}$
$⟹ s_{1} = \frac{1}{2} a_{1} t_{1}^{2} + \frac{a_{1}^{2} t_{1}^{2}}{2 a_{2}}$
$v_{1} = \frac{s_{1}}{t_{1} + t_{2}} = \frac{s_{1}}{t_{1} + \frac{a_{1}}{a_{2}} t_{1}} = \frac{1}{2} a_{1} t_{1}$
In second case,
Distance traveled during acceleration= $a_{1} t_{1}^{2}$
Distance traveled during retardation= $\frac{(2 a_{1} t_{1})^{2}}{2 a_{2}}$
$⟹ s_{2} = a_{1} t_{1}^{2} + \frac{2 a_{1}^{2} t_{1}^{2}}{a_{2}}$
$v_{2} = \frac{s^{2}}{t_{1} + t_{3}} = \frac{s_{2}}{t_{1} + \frac{2 a_{1}}{a_{2}} t_{1}} = a_{1} t_{1}$
Hence $v_{2} = 2 v_{1}$
And $2 s_{1} < s_{2} < 4 s_{1}$

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