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**Hint:**The question is based on the kinematics. In kinematics, the equations of motion are very important because they can help us in determining velocity, acceleration and displacement of a body performing the linear motion. As a train also performs a linear motion hence we can solve this problem with the help of the equations of motion.

**Complete step by step answer:**

We know that the third equation of motion is given by,

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

Where u is the initial velocity of the object, s is the displacement of the object, with acceleration a the object reaches a velocity v, which we can find using the above equation. Now the above equation can be rearranged to find out the displacement of the object and it is given as,

$s = \dfrac{{\left( {{v^2} - {u^2}} \right)}}{{2a}}$ ………. (1)

As the length of the train is ‘L’ then putting$s = L$ and $v = 3u$ in equation (1), we get,

$L = \dfrac{{\left( {{{\left( {3u} \right)}^2} - {u^2}} \right)}}{{2a}}$

$ \Rightarrow L = \dfrac{{4{u^2}}}{a}$ ……. (2)

At the midpoint of the train, the train will have covered a distance equal to half of its length. Hence, putting,$s = L/2$ in the equation (1) we get,

$\dfrac{L}{2} = \dfrac{{\left( {{v^2} - {u^2}} \right)}}{{2a}}$

From equation (2) put the value of L in the above equation, we get,

$\dfrac{{4{u^2}}}{{2a}} = \dfrac{{\left( {{v^2} - {u^2}} \right)}}{{2a}}$

$ \Rightarrow 4{u^2} = {v^2} - {u^2}$

$\therefore v = \sqrt 5 u$

The velocity with which the middle point of the train passes the same point is found to be $\sqrt 5 u$.

**Hence, we can conclude that option C is the correct answer option.**

**Note:**Consider an object performing the linear motion. If ‘u’ is the initial velocity of the object, ‘s’ is the displacement of the object, with an acceleration ‘a’ the object reaches a velocity ‘v’, in time ‘t’. There are three equations of motion in the kinematics of rigid bodies which are given below.

$v = u + at$

$s = ut + \dfrac{1}{2}a{t^2}$

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

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