Man on a Railroad Bridge

Date: 09/20/98 at 20:04:26
From: Lee Jackson
Subject: Math

A man is three-eighths of the way across a railroad bridge when he 
hears a train behind him. The train's speed is 60mph. He concludes 
correctly that he has just enough time to run to either end of the 
bridge to prevent being hit by the train. How fast can he run? 

I concluded 80mph because I multiplied eight thirds, the multiplicative 
inverse of three eighths, by 60mph. If I am not correct, please explain 
at length.

Date: 09/21/98 at 16:48:47
From: Doctor Peterson
Subject: Re: Math

Hi, Lee. This is a fascinating problem! I solved it by rather involved 
algebra first, but then I saw a much easier way. The work is not much 
harder than what you did, but the thought behind it is a lot more 
involved. You can't just blindly look for something to divide, but need 
to think through what would be proportional to what.

The key is that the man would get to either end just in time, even 
though the two ends are different distances from him. How can that be? 
Well, the train will get to the end where he started before it gets to 
the other end, so he has that much more time to get to the far end. 
The time it takes for the train to get from one end to the other will 
be the length of the bridge divided by its speed. That has to equal the 
difference between the times for the man to get to either end, each of 
which is his distance from that end, divided by his speed. This gives 
us the equation:

   Bridge length   5/8 Bridge length   3/8 Bridge length
   ------------- = ----------------- - -----------------
   Train's speed      Man's speed         Man's speed

See if you can take it from there.

It's amazing that the answer doesn't depend on the length of the bridge 
or how far away the train is. I had to add variables for those when 
I used algebra directly.

(By the way, do you know anyone who can run 80 miles per hour?)

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/