If the engine of a long train moving with constant acceleration crosses a tree with velocity u and the last compartment of the train crosses the same tree with velocity v. then the velocity with which the middle compartment crosses the same tree is

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$\frac{(v + u)}{2}$

$\sqrt{2 (u^{2} + v^{2})}$

$\sqrt{\frac{(v^{2} + u^{2})}{2}}$

$\frac{2 u v}{(u + v)}$

answer is C.

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Detailed Solution

Here, displacement will be equal to length of train l. According to third equation of motion,
$\Rightarrow v^{2} = u^{2} - 2 a s$
$\Rightarrow v^{2} = u^{2} - 2 a l$
$\Rightarrow a = \frac{v^{2} - u^{2}}{2 l}$
The length from one end of train to its mid point $= \frac{l}{2}$
$∴$ If $v_{1}$ be the speed of midpoint, then
$v_{_{1}}^{2} - u^{2} = 2 a \times \frac{l}{2}$
$\Rightarrow v_{_{1}}^{2} - u^{2} = a l$
$\Rightarrow v_{_{1}}^{2} - u^{2} = \frac{v^{2} - u^{2}}{2 l} \times l$
$\Rightarrow v_{_{1}}^{2} - u^{2} = \frac{v^{2} - u^{2}}{2}$
$v_{_{1}}^{2} = u^{2} + \frac{v^{2} - u^{2}}{2} = \frac{2 u^{2} + v^{2} - u^{2}}{2} = \frac{u^{2} + v^{2}}{2}$
$v_{1} = \sqrt{\frac{u^{2} + v^{2}}{2}}$