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# The two ends of a train moving with constant acceleration pass a certain point with velocities u & 3u. The velocity with which the middle point of the train passes the same point is:A) $2u$B) $\dfrac{3}{2}u$C) $\sqrt 5 u$D) $\sqrt {10} u$

Last updated date: 22nd Jun 2024
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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.

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.
$v = u + at$
$s = ut + \dfrac{1}{2}a{t^2}$
${v^2} = {u^2} + 2as$