Henry Ernest Dudeney/Modern Puzzles/38 - The Despatch-Rider

by : $38$

 * The Despatch-Rider
 * If an army $40$ miles long advances $40$ miles
 * while a despatch-rider gallops from the rear to the front,
 * delivers a despatch to the commanding general,
 * and returns to the rear,
 * how far has he to travel?

Solution

 * $93 \dfrac 1 3$ miles.

gives:
 * $40 + \sqrt {2 \times 40^2} \approx 96 \cdotp 568$ miles

but the author of this page can't work out why.

Proof
Let $d_1$ miles be the distance travelled by the despatch-rider from the rear to the front.

Let $d_2$ miles be the distance travelled by the despatch-rider from the front back to the rear.

Let $v_d$ be the speed of the despatch-rider.

Let $v_a$ be the speed of the army.

Let $t$ hours be the time taken for the despatch-rider to return to the rear of the army.

The despatch-rider needs to travel from the rear of where the army was to the front of where the army is now.

That is, $40$ miles plus $40$ miles.

So the despatch-rider travels $80$ miles while the army moves $40$ miles.

Hence:
 * $v_d = 2 v_a$

Then the despatch-rider travels a distance:
 * $d_2 = v_d t$

while the rear of the army travels a distance:
 * $\paren {40 - d_2} = v_a t = v_a \dfrac {d_2} {2 v_a}$

Hence:
 * $40 - d_2 = 2 d_2$

and so:
 * $d_2 = \dfrac {40} 3$

Hence the despatch-rider travels $80 + \dfrac {40} 3 = 93 \dfrac 1 3$