The problem states that a train is moving with constant acceleration and passes a certain point with velocities u and 3u at its two ends. We need to find out the velocity with which the middle point of the train passes the same point.



We can use the equations of motion to solve this problem. Let the length of the train be L, and let t be the time taken by the train to pass the given point. Then, we have:

- Velocity of the first end of the train, v1 = u + at
- Velocity of the second end of the train, v2 = 3u + at
- Distance covered by the first end of the train, s1 = ut + 1/2at^2
- Distance covered by the second end of the train, s2 = 3ut + 1/2at^2
- Distance between the two ends of the train, L = s2 - s1 = 2ut + 1/2at^2



Let the velocity of the middle point of the train be vm. Then, we have:

- Velocity of the first end of the train relative to the middle point, v1m = vm - u
- Velocity of the second end of the train relative to the middle point, v2m = 3u - vm
- Average velocity of the train, vavg = (v1 + v2)/2 = 2u + at
- Velocity of the middle point of the train, vm = vavg - 1/2a(L/2)

Substituting the values of v1 and v2 in terms of v1m and v2m, we get:

vavg = (v1m + u + v2m + 3u)/2 = v1m + 2u + v2m/2

Substituting the value of L in terms of u, a, and t, we get:

vm = vavg - (u + 1/4at^2)

Simplifying the above equation, we get:

vm = (v1m + v2m)/2

Therefore, the velocity with which the middle point of the train passes the given point is the average of the velocities of the two ends of the train relative to the middle point, which is independent of the acceleration of the train.