Aidan and Bradwick are standing back to back next to a railroad track.
When the front of a train passes them, Aidan starts to walk in the opposite direction of the train, while Bradwick starts to walk in the same direction as the train. Each stops walking when the back of the train passes him.
If the two of them walk at the same speed, and Aidan walks exactly 30 feet, and Bradwick walks exactly 45 feet, how long is the train?
I originally had a solution like Steven, but then I thought of a more clever way to reduce the amount of algebra.
Let's start when Aidan stops with the rear of the train. Then Bradwick is 60 feet away. Bradwick then walks another 15 feet before the rear of the train catches up with him. So the train travels 75 feet to Bradwick's 15 feet. Therefore the train's speed is five times faster than the guys' speed.
Now knowing the relative speeds, let's go back to the beginning of the scenario. It takes some amount of time for Bradwick to walk 45 feet. In that same amount of time the train travels those 45 feet plus its length; call the length L and then the train travels 45+L feet.
We know the train is five times faster than Bradwick. That means 5*45 = 45+L; or L=180. The train is 180 feet long.
I can easily generalize. Say Aidan walks A feet and Bradwick walks B feet. Then the train is (A+B)/(B-A) times faster then the guys. Subsequently the length of the train is found by (A+B)/(B-A)*B = B+L, which makes L=2*A*B/(B-A) as the length of the train.
A quick substitution for the specific problem: 2*30*45/(45-30) = 180.
An aside - I looked up sizes of typical railcars and got 50-60 feet per railcar and the train engine coming in around 70 feet. So the train in this problem looks like a single engine pulling two cars. Not much of a train.
Solution
