Problem:
A column of troops 3 miles long marches steadily down a straight road. An officer rides at a constant rate from the rear of the column to the head and immediately back to the rear. If the officer travels at a speed equal to twice that of the troops, how much of an advance will the troops make during one such circuit by the officer?
Answer:
To solve this problem, we’ll determine the time it takes for the officer to complete his circuit (from the rear to the head and back to the rear) and then calculate how much distance the troops cover during this time.
Definitions:
- $ L = 3 $ miles (length of the column)
- $ v $ = speed of the troops
- Officer’s speed $ = 2v $
Step 1: Time from Rear to Head (First Leg)
The officer and the head of the column are both moving forward. The officer’s relative speed towards the head is:
$$ \text{Relative speed} = \text{Officer’s speed} – \text{Troops’ speed} = 2v – v = v $$
Time taken to reach the head:
$$ t_1 = \frac{\text{Distance to head}}{\text{Relative speed}} = \frac{L}{v} $$
Step 2: Time from Head Back to Rear (Second Leg)
When the officer returns, he moves backward relative to the ground at speed $ 2v $, while the rear of the column moves forward at speed $ v $. The relative speed between the officer and the rear is:
$$ \text{Relative speed} = 2v + v = 3v $$
Time taken to return to the rear:
$$ t_2 = \frac{\text{Distance between head and rear at time } t_1}{\text{Relative speed}} $$
At time $ t_1 $, the distance between the head and rear is:
$$ \text{Distance} = x_{\text{head}}(t_1) – x_{\text{rear}}(t_1) = (L + vt_1) – (vt_1) = L $$
So,
$$ t_2 = \frac{L}{3v} $$
Step 3: Total Time and Distance Advanced by Troops
Total time $ T = t_1 + t_2 $:
$$ T = \frac{L}{v} + \frac{L}{3v} = \frac{4L}{3v} $$
Distance advanced by the troops:
$$ \text{Distance} = v \times T = v \times \frac{4L}{3v} = \frac{4L}{3} $$
Substituting $ L = 3 $ miles:
$$ \text{Distance} = \frac{4 \times 3}{3} = 4 \text{ miles} $$
Answer:
4 miles