An engine of a train, moving with uniform acceleration, passes the signal-post with velocity $$u$$ and the last compartment with velocity $$v$$. The velocity with which middle point of the train passes the signal post is:

Let the length of the train be $$L$$ and the uniform acceleration be $$a$$. The front of the train (engine) passes the signal post with velocity $$u$$ and the last compartment passes with velocity $$v$$. Using the kinematic relation $$v^2 = u^2 + 2aL$$, we get $$a = \frac{v^2 - u^2}{2L}$$.

The middle point of the train is at a distance $$\frac{L}{2}$$ from the engine. When the middle point passes the signal post, the engine has travelled a distance $$\frac{L}{2}$$ beyond the post. Let the velocity of the middle point at the signal post be $$v_m$$. Using the kinematic equation for the engine travelling from the signal post to a distance $$\frac{L}{2}$$:

$$v_m^2 = u^2 + 2a \cdot \frac{L}{2} = u^2 + aL$$

Substituting $$aL = \frac{v^2 - u^2}{2}$$:

$$v_m^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$$

Therefore $$v_m = \sqrt{\frac{u^2 + v^2}{2}}$$, which corresponds to Option (2).