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 (1) 2u (2) 2 3 u (3) 5 u (4) 10

Answer 1

Let L be the total length of train

acceleration Be "a" m/s

Given that when starting point of train passes through the point, Speed = "u" m/s

At the end Trains velocity = 3u m/s

Formula used :- V2 - U2 = 2as ( v square and u square)

Substituting train velocities in given formula we get

9u2 - u2 =2aL

a = 4u2 /L

Now velocity passed ag s = L/2

v2-u2 = 2.a.L/2

v2=u2+L.4u2/L

v2=5u2

v = 5u