A Close Call

Here is a problem by J.A.H. Hunter.

Returning from a tiring routine of inspection, Bob had his track motor three-eighths of the way across that long single track bridge when his lights failed. And it was just then that he heard the distant thunder of the approaching train. Instinctively, he knew it was the Glen Flyer, scheduled to be doing a steady 60 mph at that point; and Bob headed right back at full speed. Little remained of his track motor after the collision, but Bob had just time to jump clear and roll safely down the embankment at the end of the bridge where where the crash occurred. The subsequent inquiry proved his instinct had been right. If he had tried to escape the other way, the train would have caught him at the far end of the bridge, but there he would have jumped to certain death. So what was the speed of his track motor?

Please give the answer and the way to get the answer.

Peter Heichelheim

In order for this puzzle to make sense, the train has to be coming from behind Bob.

t = amount of time it takes Bob to cross 3/8 of the bridge = amount of time for the train to get from where Bob heard it to the leading edge of the bridge

If Bob had gone the other way instead, i.e., continued crossing the bridge, he would have gone across 5/8 of the bridge since hearing the train. In that amount of time, the train (and Bob) would have gone 5/3 longer than they actually went.

Defining bridge = length of bridge, then the time for Bob to cover 3/8 of the bridge is 3/8*bridge / (Bob's speed). In that time, the train covers 60mph * (3/8 * bridge / (Bob's speed)).

60mph * 3/8 bridge / Bob's speed = distance from train when Bob heard it to the leading edge of the bridge.

60 mph * 5/8 bridge / Bob's speed = distance from train when Bob heard it to the leading edge of the bridge _plus_ the length of the bridge.

Subtracting the two equations, we get:

bridge = 1/4 * bridge * 60 mph / Bob's speed

1 = 1/4 * 60 mph / Bob's speed

Bob's speed = 15 mph.

-Doug Magnoli [Delete the two and the three for email.]

When the train hits Bob's motor, Charlie still has a distance of 2x to cover and the train is a distance 8x behind him moving at 60mph. So if they meet at the end of the bridge, the motor must travel at 2/8 the speed of the train.

You and Mike Williams got J.A.H. Hunter's answer which is 15 mph. I particularly like the way you used to solve it. If he was facing the train (I originally thought he was doing this) there would be no question which way Bob would go as he could not make both ends of the bridge with the train hitting him at both places. This problem is from May 1956, but Bob would seem to have a pretty dangerous job. I can imagine if this routine is permitted there would be quite a few fatal accidents.