perplexus.info :: General : Stay in Form

Stay in Form (Posted on 2004-01-03)

A body of soldiers form a 50m-by-50m square ABCD on the parade ground.

In a unit of time, they march forward 50m in formation to take up the position DCEF.

The army's mascot, a small dog, is standing next to its handler at location A. When the soldiers start marching, the dog begins to run around the moving body in a clockwise direction, keeping as close to it as possible. When one unit of time has elapsed, the dog has made one complete circuit and has got back to its handler, who is now at location D (assume the dog runs at a constant speed and does not delay when turning the corners).

 B----C----E
 |    |    |   
 A----D----F

How far does the dog travel?

Submitted by DJ

Rating: 4.2500 (8 votes)

Solution: (Hide)

Let L be the side of the square, 50m.
Let D be the distance the dog travels.
Let T be the total time it took the soldierd to march 50m.
Let v1 be the soldiers' marching speed and v2 be the speed of the dog.

Then, v1 = L/T and v2 = v1*D/L = D/T.

Let t1, t2, t3, t4 be the time the dog takes to traverse each side of the square, in order. Find t1 through t4 in terms of L and D and solve t1+t2+t3+t4 = T:

While the dog runs along the back edge of the square in time t1, the soldiers advance a distance d=t1*v1, so the dog has to cover a distance:

√(L² + (t1*v1)²)
This takes a time:

t1=√(L² + (t1*v1)²)/v2 . . .
t1 = L/√(v2² - v1²).
Also, t3 is the same path, simply in the other direction, so:

t3 = t1 = L/√(v2² - v1²).
Simpler to come up with are:
t2 = L/(v2-v1) t4 = L/(v2+v1)
Therefore, we have:
t1 + t2 + t3 + t4 = T 2L/√(v2² - v1²) + L/(v2-v1) + L/(v2+v1) = T 2L/√[(D/T)²-(L/T)²] + L/[(D/T)-(L/T)] + L/[(D/T)+(L/T)] = T 2L/{[√(D²-L²)]/T} + L/[(D-L)/T] + L/[(D+L)/T] = T 2L/[√(D²-L²)] + L/(D-L) + L/(D+L) = 1 2L/[√(D²-L²)] + 2DL/(D²-L²) = 1 2L[D + √(D²-L²)]/(D² - L²) = 1 . . . D^4 - 4LD^3 - 2L^2D^2 + 4L^3D + 5L^4 = 0
This last equation has a root D = 4.18113L, or D = 209.056m when L=50m.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
answer	K Sengupta	2008-02-07 05:15:52
im not too bright	Billy Bob	2004-03-19 16:54:15
solution	luminita	2004-01-05 05:20:17
The answer is.........	sunny	2004-01-04 17:59:50
No Subject	GT	2004-01-03 15:49:47
I get...	Charlie	2004-01-03 14:58:17
re(2): i think this is the answer	Victor Zapana	2004-01-03 14:03:25
re: i think this is the answer	rerun141	2004-01-03 13:36:07
i think this is the answer	Victor Zapana	2004-01-03 13:13:54

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