A train is moving forward at a velocity of 2.0m/s. At the instant the train begins to accelerate at 0.80 m/s² a passenger drops a coin which takes 0.50s to fall to the floor. Relative to a spot on the floor directly under the coin at release, it lands.

When the passenger drops the coin, it will have an initial velocity of 2m/s in the horizontal direction(that of the train).

The horizontal range of the coin, during the given time of descent, is 
R = ut
 = (2m/s)(0.5s)
 = 1m

The distance traveled by train in the same time is given by
S = ut+1/2at²
 = 1 + (0.5)(0.8m/s²)(0.5s)²
 = 1.1 m

The train moved 1.1m while the coin move only 1 m horizontally in the time of flight. Therefore, the coin will be 0.1m to the rear of the train.