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A column of troops one km long is moving along a straight road at a uniform pace. A messenger is sent from the head of the column, delivers a message at the rear of the column and returns. He also moves at a uniform pace and arrives back at the head of the column when it has just covered its own length. How far did the messenger travel?

I can't get any ideas on how to start.

Thanks for any help.

Matt
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  • How far did the messenger travel means the distance traveled or the displacement? If its displacement then the answer should be 1km. – Saurabh Jul 03 '12 at 06:10
  • It means distance. – Matt Jul 03 '12 at 06:12
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    You just need to organise the information carefully and use distance = speed x time. Write down what you know about the movement of the column - say speed $u$ and the messenger say speed $v$ taking total time $t$. Split the movement of the messenger into two obvious parts. Then identify the total distance $d$ you want to find, and use the information you have to find an equation for $d$. – Mark Bennet Jul 03 '12 at 06:19
  • A trivial answer to the question would be $1$km, the length of the column. After all, the messenger starts out at the head of the column, and ends up again at the end of the column, at the point in time when it has covered its own length, and she then has advanced as much as the column has. This is probably not what the question intends, but it could be made clearer. – Marc van Leeuwen Jul 03 '12 at 06:50

4 Answers4

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Let us assume that the speed of the column is $1$ km per unit of time. For convenience, call that unit an hour. The column took $1$ hour to cover its own length.

Let $v$ be the speed of the messenger. When she is travelling to the back, the combined speed of approach of the messenger and the column rear is $v+1$, so the time it takes is $\frac{1}{v+1}$. Going the other way, the speed at which the messenger gains on the head is $v-1$, so the time it takes to gain the whole $1$ km is $\frac{1}{v-1}$. The whole task took $1$ hour, and therefore $$\frac{1}{v+1}+\frac{1}{v-1}=1.$$ This gives $v=1+\sqrt{2}$. The time taken is an hour, so the distance travelled is $1+\sqrt{2}$.

André Nicolas
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Let the speed of troop is $u$ kmph and that of messenger is $v$ kmph.
So the relative speed when messenger is going backward is $v+u$ kmph.
And relative speed while going forward is $v-u$ kmph $(v>u)$.
So the total time taken is $$t=\frac{1}{v+u}+\frac{1}{v-u}$$ but in this time troop had moved $1$km so the $t=\frac{1}{u}$
$$\frac{1}{u}=\frac{1}{v+u}+\frac{1}{v-u}$$ $$v^2-2uv-u^2=0$$ which give us the following relation $$v=u(1+\sqrt2)$$

so the distance traveled is $$\begin{align*} d= &\underbrace{\frac{v}{v+u}}_{backward}+\underbrace{\frac{v}{v-u}}_{forward}\\ &=\frac{2v^2}{v^2-u^2}\\ &=\frac{2v^2}{2uv}\\ &=\frac{v}{u}=1+\sqrt{2}\text{ km}\end{align*}$$

Saurabh
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  • Mostly ok, but why did you give the answer using units of speed? Mind you, I've never seen it abbreviated "kmph". At least locally "km/h" is the only way to write it. This is different from, e.g. U.S.A., where "mph" is always used. I guess there are local variations :-) – Jyrki Lahtonen Jul 03 '12 at 07:27
  • @JyrkiLahtonen Oh that was a mistake .I have corrected it.Thanks – Saurabh Jul 03 '12 at 07:32
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Let $x$ be the troops' speed and $y$ be the messenger's speed.

Total messenger travelling time is $1/x$ (as the troops moved 1km forward).

In the troops frame of reference, the messenger first moved backwards with the speed of $y+x$, and then moved forward with the speed of $y-x$ so that in the $1/x$ total time he returned to the initial position.

Let $z$ be the time messenger spent to reach the rear of the column. Obviously, $z = 1/(y+x)$ (as messenger moves with the speed of $y+x$ relative to the column and has to travel $1$ km relative to the column to reach its rear).

From the other side, we have $z \times (y+x) + (1/x - z) \times (y-x) = 0$, from which follows $2zx + y/x - 1 = 2/(y/x+1) + y/x - 1 = 0$.

Note that the total messenger travelling distance is $d = y/x$. From the previous equation we get $2/(d+1) + d - 1 = 0$, $d^2-3 = 0$, and thus $d = \sqrt{3}$.

penartur
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Since the time the column travels is equal to the time the messenger travels, then D= rt makes the distance equal to the rate if we let t = 1. Suppose the distance the column moves until the messenger overtakes it is x, then the distance the messenger travels is 1+x, making the comparison 1+x/ x. On the messenger's return, the distance is x while the column moves 1-x. So: 1+x/x = x/1-x since both column and messenger move at constant rates. x then is equal to 1/2 the square root of 2, so the messenger travels 1+ half the square root of 2 out and another half the square root of two back. Total distance of 1 plus the square root of two.

William Pauch
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