Q. A particle is moving with constant acceleration from $A$ to $B$ in a straight line $AB$. If $u$ and $v$ are the velocities at $A$ and $B$ respectively then its velocity at the midpoint $C$ will be
A) $\left( \frac{u^{2} + v^{2}}{2u} \right)^{2}$
B) $\frac{u + v}{2}$
C) $\frac{v - u}{2}$
D) $\sqrt{\frac{u^{2} + v^{2}}{2}}$
Explanation
(d) [image:3_image196.emf]
Let $S$ be the distance between $AB$ and $a$ be constant acceleration of a particle. Then
$v^{2} - u^{2} = 2aS$
$Or$ $aS = \frac{v^{2} - u^{2}}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Let $v_{c}$ be velocity of a particle at midpoint $C$
$\therefore\ v_{c}^{2} - u^{2} = 2a\left( \frac{S}{2} \right)$
$v_{c}^{2} = u^{2} + aS = u^{2} + \frac{v^{2} - u^{2}}{2}$ [Using (i)]
$v_{c} = \sqrt{\frac{u^{2} + v^{2}}{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.