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

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

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

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

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

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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)

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

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

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

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

Watch 3-min video & get full concept clarity