1. Set up the motion

Let the length of the train be $L$ and its uniform acceleration be $a$.

When the engine passes the signal post, the train's speed is $u$. When the last compartment passes the signal post, the train's speed is $v$.

This means that during the time from engine crossing the post to last compartment crossing the post, the train has moved forward by a distance equal to its own length $L$.

So, using the equation of motion,

$v^2 = u^2 + 2aL \, .$

Hence,

$2aL = v^2 - u^2.$

2. Consider the instant when the middle point passes the signal post

When the middle point of the train passes the signal post, the train has moved by a distance

$\frac{L}{2}$

from the instant when the engine passed the signal post.

Let the speed at that instant be $V$.

Again using

$V^2 = u^2 + 2a\left(\frac{L}{2}\right),$

we get

$V^2 = u^2 + aL.$

From earlier,

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

Substitute this:

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

Simplify:

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

Therefore,

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

3. Match with the options

This corresponds to:

$\boxed{\sqrt{\frac{v^2 + u^2}{2}}}$

which is Option D.

4. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.