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).
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How far does the dog travel?
If s is the speed of the dog relative to that of the troop, the time taken would be twice the time taken to go from side to side plus the time taken to get from the trailing edge to the leading edge plus the time taken to get from the head to the trailing edge of the troop:
2/√(s²-1) + 1/(s-1) + 1/(s+1)
This is to be completed in unit time.
Excel's solver makes s be equal to 4.18112545 in order for this to be 1. As the troop is traveling at 50 m per unit of time, the dog is going 209.056272 m
per unit of time, and that's the distance he's traveled.
That includes two crosswise traverses taking .246318793 units of time each, during which he travels 51.49449 m each, and the troops had moved 12.31594 m each time.
The forward travel took .314354155 units of time, during which the dog traveled 65.71771 m and the troops 15.71771 m.
The backward travel took .193008259 units of time during which the dog traveled 40.34959 m and the troops 9.650413 m.
Edited on January 3, 2004, 4:27 pm
Posted by Charlie
on 2004-01-03 14:58:17