Question

A particle moving with a constant acceleration from A to B in the straight line AB has velocities u and v at A and B respectively. If C is the mid-point of AB then the velocity of particle while passing C will be:

• A. $\sqrt{\frac{v^2+u^2}{2}}$
• B. $\frac{v+u}{2}$
• C. $\frac{v-u}{2}$
• D. $\frac{\left[\frac{1}{v}+\frac{1}{u}\right]}{2}$

Answer 1

The correct option is A $\sqrt{\frac{v^2+u^2}{2}}$

Let the distance between A and B=$s$ and as mentioned C is the midpoint of A and B thus distance between A C and

C B$=\frac{s}{2}$

From equation of motion we get distance between AB

$v^2=u^2+2as$

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

Let speed of the vehicle be $pm/s$ at point C

Again from equation of motion we get,

$p^2=u^2+2a\frac{s}{2}$

as distance between A and C is $\frac{s}{2}$

$s=\frac{p^2-u^2}{a}$-------(2)

Computing 1 and 2 we get

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

$p^2=\frac{v^2{+u}^2}{2}p=\sqrt{\frac{v^2{+u}^2}{2}}$