## Task 6 - 130736

A train traveling at a constant speed took exactly 27 s to cross a 225 m long bridge (calculated from the moment the locomotive entered the bridge until the last car left the bridge).

The train passed a pedestrian walking in the opposite direction of the train in exactly 9 s. During this time, the pedestrian had walked exactly 9 m.

Determine the length of the train (in meters) and its speed (in kilometers per hour)!
One could imagine that the pedestrian leaves the bridge at the same moment the locomotive enters the bridge, moving in the opposite direction to the train. If the pedestrian has covered 9 m after 9 s, the last car of the train passes by him. The last car still needs 18 s to leave the bridge because of \(27 - 9 = 18\). During this time, it covers exactly 234 m because of \(225 + 9 = 234\).

Consequently, the average speed of the train is \(234 \div 18 = 13 \, \text{m/s}\), which is \(13 \times \frac{3600}{1000} = 46.8 \, \text{km/h}\).

Since the train took 27 s to cross the bridge, the locomotive covered \(13 \times 27 = 351 \, \text{m}\) during this time. This distance consists of the 225 m length of the bridge and the length of the train. Therefore, the length of the train is \(351 - 225 = 126 \, \text{m}\).

Solutions of the III. Round 1973 adopted from [5]

### 4.16 XIV. Olympiad 1974

### 4.16.1 I. Round 1974, Class 7