An army is marching in single file in a line that stretches

2

21385

New member
Joined
Dec 18, 2006
Messages
7
An army is marching in single file in a line that stretches 20 km. A dispatch rider travels from the rear to the front, delivers a message, and returns to the rear. In the time the dispatch rider takes to do this, the army advances 20 km. How far did the dispatch rider travel in going from the rear to the front and back to the rear again. (note that the rider's turnaround time at the front is instantaneous)

Please show solution. thanks
 
Let D=the distance the army marches

Then, 20+D=the distance the rider travels during trip to the front.

20-D=the distance traveled on the way back.

Then you have:

\(\displaystyle \L\\\frac{20+D}{D}=\frac{D}{20-D}\)

Solve for D.

The entire trip he will have traveled 20+2D km.


I hope I didn't wonder off into the darkness on this one.
 
i still don't understand how you did it...can you explain a bit more? thanks
 
i know the answer is 20+20sqrt(2) but i need to understand the concept. thanks
 
The equation is due to the fact that the ratio between the distance the

army marches and the distance the rider travels must be the same both

ways. Picture the rider starting out from the rear and heading to the

front of the line. The army is marching this whole time. The army is 20

km long plus the distance travelled, 20+D.

The way back would be 20-D.

The entre trip would be 20+D+D=20+2D.

Solve the equation from the last post for D.
 
Here's a way to look at what galactus did....the long way:

[A]X......20......Y......20......[C]

Since army moves 20 km, we can simplify with
2 guys X and Y, X being the rider, Y the army's
front; X catches up to Y somewhere within BC;
X then goes back to B, Y continues to C, both
arriving at same time.

x = X's speed, y = Y's speed, k = distance travelled by Y when X catches up to him.

X catches up to Y:
(k + 20) / x = k / y
k(x - y) / y = 20 [1]

X back to B while Y goes to C:
k / x = (20 - k) / y
k(x + y) / x = 20 [2]

Combining [1],[2]; k's cancel out and:
y(x + y) = x(x - y)
y^2 + 2xy - x^2 = 0 ; using quadratic:
y = x[sqrt(2) - 1]

Substitute in [2]:
k[x + x(sqrt(2) - 1)] / x = 20 ; x's cancel out:
k + k[sqrt(2) - 1] = 20
k = 20 / sqrt(2)

So X travelled 20 + 2k = 20 + 40/sqrt(2) = 20 + 20sqrt(2)
 
You must log in or register to reply here.
Top