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Question: The two ends of a train moving with constant acceleration pass a certain point with velocities u & 3...

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) 2u2u
B) 32u\dfrac{3}{2}u
C) 5u\sqrt 5 u
D) 10u\sqrt {10} u

Explanation

Solution

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,
v2=u2+2as{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=(v2u2)2as = \dfrac{{\left( {{v^2} - {u^2}} \right)}}{{2a}} ………. (1)
As the length of the train is ‘L’ then puttings=Ls = L and v=3uv = 3u in equation (1), we get,
L=((3u)2u2)2aL = \dfrac{{\left( {{{\left( {3u} \right)}^2} - {u^2}} \right)}}{{2a}}
L=4u2a\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/2s = L/2 in the equation (1) we get,
L2=(v2u2)2a\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,
4u22a=(v2u2)2a\dfrac{{4{u^2}}}{{2a}} = \dfrac{{\left( {{v^2} - {u^2}} \right)}}{{2a}}
4u2=v2u2\Rightarrow 4{u^2} = {v^2} - {u^2}
v=5u\therefore v = \sqrt 5 u
The velocity with which the middle point of the train passes the same point is found to be 5u\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+atv = u + at
s=ut+12at2s = ut + \dfrac{1}{2}a{t^2}
v2=u2+2as{v^2} = {u^2} + 2as