The Ratio of the Speed of a Colonel to his Soldiers

Date: 11/7/95 at 22:31:4
From: Charles Sismondo
Subject: Is our answer correct?

A colonel leads a column of soldiers.  He marches to the rear of the 
column at a speed different from the column's and then at the same speed 
marches back to the front of the column.  During this time the column 
marches a distance equal to its length.  What is the ratio of the speed 
of the faster to the slower?  Our answer, intuitively, is the colonel is 
faster and the ratio is 2:1.

Is that correct, and why?

Thanks.

Date: 11/10/95 at 11:43:49
From: Doctor Ken
Subject: Re: Is our answer correct?

Hello there!

I think it's worth your while to look at the following picture:

 |
 |                        /
 |                      //
 |                    / /
 |                  /  /
 |         troops /   /
d|              /    /
i|            /     /
s|          /      /
t|        /       /
a|      /        /
n|       \      /
c|        \    /
e|         \  /
 |          \/
 |        colonel
 |  
 |_____________________________________
            time

Does that make sense?  Then you have to find out just where that "v" 
point is on the graph of the colonel's march.  It's determined by where 
the back of the column is, right?  So you can find the relative slopes 
of the two marching folks by figuring out where the tip of the colonel's 
"v" is.  Then you can find out whether your intuition was correct.

Good luck!

-Doctor Ken,  The Geometry Forum

Date: 9/12/97 17:59:48
From: Max 
Subject: Problem for a math contest

The following is a question from a math contest, I have tried drawing 
diagrams and pictures but it just really confuses me. I don't know 
where to start. Here goes:

On a straight road, an inspecting officer traveled from the rear 
to the front of an army column and back while the column 
marched forward its own length. If the officer and the column
maintained steady (but different) speeds, what was the ratio of 
their speeds, faster to slower?

Date: 9/13/97 13:07:00
Subject: Re: Problem for a math contest

To answer this sort of problem you must introduce letters to represent 
quantities you don't know. You can then write down equations to 
represent the information given in the question.

Let u = the speed of the column
    v = the speed of the officer
    L = the length of the column

The time needed for the column to march its own length = L/u

Now impose a velocity -u on the whole system. This brings the column to 
rest and the officer moves at v-u in going from rear to the front, and at 
v+u in going from front to rear.

Because the column is at rest (in our new setup) the distance travelled by 
the officer is L in each direction. The time for the officer to travel both 
ways is:

     L/(v-u) + L/(v+u) =  L/u          divide both sides by L
     
           v+u + v-u        1
           ----------  =   ---
           v^2 - u^2        u

                2v          1
            ---------  =   ---
            v^2 - u^2       u

                   2vu =   v^2 - u^2   divide out by u^2

                2(v/u) =  (v/u)^2 - 1  so we have a quadratic in (v/u)

  (v/u)^2 - 2(v/u) - 1 = 0

                           2 +-sqrt(4 + 4)
                 (v/u) =  ----------------
                                 2

                           2 +-2sqrt(2)
                       =   ------------
                                2

                 (v/u) =  1 +-sqrt(2)    we cannot have v/u <0 
                                         so take positive sign.
                 (v/u) = 1+sqrt(2)


    Speed of officer
    ----------------   =   1 + sqrt(2)
    Speed of column
      

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/