Two ends of a train moving with a constant acceleration pass a certain point with velcities ` u and v`. Show that the velocity with which the middle point of the train passes the same point is `sqrt(9u"^(2) +v^(2)//2).

Explanation

Let (x) be the total length of the train, (V) be the velocity of the train while passing a certain middle point and (a) be the uniform acceleration of the train , Taking the motion of the train when middle point is passing from the given point, we have
` u=u, V, S=//2 , a=a`
Using, ` `v^(2) =u^(2) + 2 aS, we have
` V&(2) =u ^(2) + 2 ax//2 =u + ax` ....(i)
Taking the motion of train when the last end of train is passing from the given point, the
`u=u, v, a=a, S=x`
Now, we have,
`v^(2) =u^(2) ax or ax (v^(2) -u^(2))/2`
Putting this value in (i), we get
` V^(2) =u^(2) + (v^(2)-v^(2))/2 =(u^(2)+ v^(2))/2`
or `V=sqrt ((u^(2) + v^(2))//2)`.