Man and Train on a Bridge

Date: 2/1/96 at 16:27:56
From: Anonymous
Subject: Story problem

A man is walking across a railroad bridge that goes from point A 
to point B. He starts at point A, and when he is 3/8 of the way 
across the bridge, he hears a train approaching. The train's speed 
is 60 mph (miles per hour).  

The man can run fast enough so that if he turns and runs back 
toward point A, he will meet the train at A, and if he runs 
forward toward point B, the train will overtake him at B.  

How fast can the man run?

I have no idea how to do this problem.

Date: 3/24/96 at 22:54:16
From: Doctor Rachel
Subject: Re: Story problem

Hi!

   

If the man turns and runs toward point A, he will cover 
3/8 of the length of the bridge in the time that it takes 
the train to reach A. 

If the man runs forward toward point B, what part of the bridge 
will he cover before the train reaches A? Well, he will cover 
3/8 of the bridge, only heading forward toward B. This will put 
him 3/8 + 3/8 = 6/8 = 3/4 of the way across the bridge by the 
time the train reaches A.

Since we know that the man and the train will meet at B, this 
means that in the time it takes the man to run the remaining 
1/4 of the bridge, the train will cover the entire length of 
the bridge.

If it takes the man the same time to cover 1/4 of the bridge 
that it takes the train to cover the whole bridge, then the train 
must be going four times as fast as the man. Another way of saying 
this is that the man runs at 1/4 the speed of the train.  

Since the train's speed is known to be 60 mph, this means that 
the man runs at (1/4) 60 = 15 mph.
_____________________________________________________________

An equation-oriented solution to this problem makes use of the 
distance formula, which tells us that time = distance/rate.  
We will use this to express the information that we know about 
the train and the man when the man is 3/4 of the way across the 
bridge and the train is at A.

The period of time that we are considering is the same for both 
the train and the man. During this time, however, they travel 
different distances. If the bridge has length d, the train 
travels d while the man only travels 1/4 d.  

The man's rate is unknown (it's what we want to figure out) so 
we will call it r. The train's rate is 60 mph. This information 
makes it possible to write the two following equations:

  Time it takes the man to run 1/4 the bridge length = (1/4 d)/r
  Time it takes the train to cover the entire bridge = d/60

We can set these two equations equal to each other because the 
two times are equal.  This produces the following:

     1/4 d    d
     ----- = ---
       r     60

Cross-multiplying results in:

     60 (1/4 d) = dr

We can divide both sides of the equation by d because d cannot 
possibly be zero (the bridge has a real length):

     60 (1/4) = r

which leaves us with r = 15 mph, the same answer as above.

-Doctors Judy and Rachel, The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 8/1/97 at 12:00:00 
From: Anonymous
Subject: Man walking across bridge.

I read with interest your solution to FSarre6399@aol.com 2/1/96 
Man walking across bridge problem. 

I think my solution is simpler and more elegant than the one you 
gave!

Solution:  Let M stand for the man's speed in mph.  When the man 
runs toward point A, the relative speed of the train with respect 
to the man is the train's speed plus the man's speed (60 + M).  
When he runs toward point B, the relative speed of the train is the 
train's speed minus the man's speed (60 - M).

When he runs toward the train the distance he covers is 3 units.  
When he runs in the direction of the train the distance he covers 
is 5 units. We can now write that the ratio of the relative speed 
of the train when he is running toward point A to the relative speed 
of the train when he is running toward point B, is equal to the 
inverse ratio of the two distance units or

               (60 + M)          5
                -----------  =  ---
               (60 - M)          3

              3(60 + M) = 5(60 - M)
              3*60 + 3M = 5*60 - 5M
                     8M = 120
                      M = 15 mph