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

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

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

answered Jun 26, 2013 by
edited Jan 25, 2014 by meena.p