\[(a)\;\frac{v^2}{a}\quad (b)\;\frac{v^2}{2a} \quad (c)\;\frac{2v^2}{a} \quad (d)\frac{v^2}{4a}\]

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Let x be distance between particles after t seconds Then

I particle $x_1=vt$

II particle $x_2=\large\frac{1}{2}$$at^2$

$x=x_1-x_2$

$\quad=vt-\large\frac{1}{2} $$at^2$-----(1)

For x maximum

$\large\frac{dx}{dt}$$=0=>v-at=0$

$=>t=\large\frac{v}{a}$

Substituting value of t in (1)

$x=v \bigg(\large\frac{v}{a}\bigg)-\large\frac{1}{2} a \bigg(\frac{v}{a}\bigg)^2$

$\quad=\large\frac{v^2}{2a}$

Hence b is the correct answer

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