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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 from rest. During the time that elapses before the second catches the first, the greatest distance between the particle is

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

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

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

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

answer is B.

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Detailed Solution

Let x be the distance between the particles after t seconds. Then

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

For x to be maximum

$\frac{dx}{dt} = 0$

or v-at = 0

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

substituting this value in (1) we get

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

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

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