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

Motion in a Straight Line

Solution:

Let the length of train is s, then by third equation of motion, $v^2 = u^2 + 2a \times s$ ....(1)
Where v is final velocity after travelling a distance s with an acceleration a & u is initial velocity as per question
Let velocity of middle point of train at same point is v', then
$(v')^2 = u^2 + 2a \times (s/2)$ ....(2)
By equation (1) & (2), we get
$v' = \sqrt{\frac{v^2 + u^2}{2}}$