Two trains, A and B, leave their respective starting points at the same time and travel in opposite directions. They travel at constant speeds, and pass at point M. One travels at twice the speed of the other. If one of the trains leaves five minutes late they pass at a point 2 miles from point M.

What is the speed of the slow train, in miles per hour?

Train A travels at x mph, and B travels at 2x. When they leave at the same time and travel for time t, they meet at point M.

If A leaves 1/12 hour (five minutes) late, then at time t, B is at point M and A is x/12 miles away. Since B travels twice as fast as A, they will meet at a point x/12 * 2/3 miles away, which is 2 miles.

2x/36=2
x=36 mph.

If, instead, B leaves 1/12 hour late, then at time t, A is at point M and B is 2x/12 miles away. Since A moves half as fast as B, they will meet 2x/12 * 1/3 miles
away.

The math works out the same as before, so regardless of which train leaves five minutes late, the slow train travels 36 mph.

If miles per hour had not been specified, the answer in alternate units includes 78.795 microparsecs/millennium, or 96,768 furlongs/fortnight :)

Posted by Bryan on 2003-05-10 13:21:54