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Two particles start moving from the same point along the same straight line. The first moves with constant velocity v and the second with constant acceleration a. During the time that elapses before the second catches the first, the greatest distance between the particles is

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\frac{v^{2}}{2a}

\frac{{2v}^{2}}{a}

\frac{v^{2}}{a}

\frac{v^{2}}{4a}

answer is B.

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Detailed Solution

Let x be distance between particles after ‘t’ seconds. First particle

x_{1} = vt

Second particle $x_{2} = \frac{1}{2} {at}^{2}$

$x = x_{1} - x_{2}$

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

For a maximum

$\frac{d x}{dt} = 0$

$v - at = 0$

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

Substituting value of t in (1)

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

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

$= \frac{v^{2}}{a} - \frac{v^{2}}{2a} = \frac{2 v^{2} - v^{2}}{2 a}$

$x = \frac{v^{2}}{2a}$

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