Question

The two ends of a train moving with constant acceleration pass a certain point with velocities u and 3u. The velocity with which the middle point of the train passes the same point is

Answer 1

S = (v^2 - u^2) / (2a)

If Length of train is L then
L = ((3u)^2 - u^2) / (2a)
L = 4u^2 / a —-(1)

Midpoint is at Distance L/2
S = (v^2 - u^2) / (2a)
L/2 = (v^2 - u^2) / (2a)
From equation (1)
2u^2 / a = (v^2 - u^2) / (2a)
4u^2 = v^2 - u^2
v^2 = 5u^2
v = u√5

Velocity with which middle point passes the same point is u√5.

Answer 2

Answer:

$for \: such \: midpoint \: questions \: there \: is \: a \: formulae \\ = \frac{ \sqrt{ {u}^{2} + {v}^{2} } }{2} \\ so \: the \: answer \: will \: be \\ \frac{ \sqrt{ {(3u)}^{2} + {(u)}^{2} } }{2} \\ \frac{ \sqrt{9 {u}^{2} + {u}^{2} } }{2} \\ \frac{ \sqrt{10 {u}^{2} } }{2} \\ \sqrt{5} u \\ so \: the \: answer \: is \: \sqrt{5} u$

Explanation:

hope this will help you