A particle is moving with constant acceleration from A to B in a straight line AB. If U and V are the velocities of particle at A and B respectively, then its velocity of particle at the midpoint C will be

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${(\frac{U^{2} + V^{2}}{2})}^{2}$

$\frac{U^{2} + V^{2}}{2}$

$\frac{V - U}{2}$

$\sqrt{\frac{U^{2} + V^{2}}{2}}$

answer is D.

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Detailed Solution

$l e t c b e t h e m i d p o i n t o f A B, l e t v e l o c i t y a t c b e x f r o m A t o C; i n i t i a l v e l o c i t y = u; f i n a l v e l o c i t y = x; d i s \tan c e = \frac{d}{2}; a c c e l e r a t i o n = a s u b s t i t u t e i n v^{2} = u^{2} + 2 a s - - - (1) X^{2} = U^{2} + 2 a \cdot \frac{d}{2} \to -- (2) f r o m A t o B; i n i t i a l v e l o c i t y = u; f i n a l v e l o c i t y = v; d i s \tan c e = d; a c c e l e r a t i o n = a s u b s t i t u t e i n e q u a t i o n (1) v^{2} = u^{2} + 2 a d \Rightarrow a d = \frac{v^{2} - u^{2}}{2} - - - (3) s u b s t i t u t e e q u a t i o n (3) i n (2) X^{2} = U^{2} + \frac{v^{2} - u^{2}}{2} X = \sqrt{\frac{v^{2} + u^{2}}{2}} = v e l o c i t y o f t h e p a r t i c l e a t m i d p o i n t o f A B$