# 10 - Messenger Going From Front To Rear Then To Front Of A Marching Army Column

**Problem**

An army of troops is marching along a road at 5 kph. A messenger on horseback was sent from the front to the rear of the column and returns immediately back. The total time taken being 10 minutes. Assuming the messenger rides at the rate of 10 kph, determine the length of the column.

**Answer Key**

[ 625 m ]

**Solution**

Let

relative velocity = 10 + 5 = 15 kph

relative distance traveled =
relative velocity = 10 - 5 = 5 kph

relative distance traveled =

*x*= length of the army column in kmGoing from front to rear:

relative distance traveled =

*x*km

Going from rear to front:

relative distance traveled =

*x*km

Note: time = distance / velocity

$t_\text{front to rear} + t_\text{rear to front} = 10 ~ \text{minutes}$

$\dfrac{x}{15} + \dfrac{x}{5} = \dfrac{10}{60}$

$\dfrac{4}{15}x = \dfrac{1}{6}$

$x = 0.625 ~ \text{km}$

$x = 625 ~ \text{m}$ ← *answer*

Tags: