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Topic: Marching Army (I) (Read 1021 times) |
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ThudnBlunder
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Marching Army (I)
« on: May 6th, 2003, 9:35pm » |
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An army one mile long begins to march in single file at a uniform rate. It marches for exactly one mile and then stops. At the same moment that the army starts marching, a messenger who is at the rear also starts to run at a uniform rate. He runs alongside just fast enough to reach the front, turn around instantaneously, and return to the rear again at the exact instant that the army stops. How far did the messenger run?
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| « Last Edit: May 6th, 2003, 9:37pm by ThudnBlunder » |
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towr
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Re: Marching Army (I)
« Reply #1 on: May 7th, 2003, 12:29am » |
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let's say the army has speed a, and the messenger speed m
now the first part the messenger has to run a mile with the relative speed m-a, the second part he has to run a mile with speed a+m. This gives you the amount of time he walks and gives
1/(m-a) + 1/(m+a) which is 1 unit of time, the same time the army has for walking the mile
so also a=1, and we can now solve
1/(m-1) + 1/(m+1) = 1
so m = 1 - sqrt(2) or m = sqrt(2) + 1, the latter of which is the answer, since a distance can't be negative.
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wowbagger
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Re: Marching Army (I)
« Reply #2 on: May 7th, 2003, 2:22am » |
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I'll have to disagree on your answer, towr.
The messenger runs "at a uniform rate", not with two different speeds on his way to the front and back.
So I come up with two miles for the distance he runs. Just consider the frame of reference that moves along with the army.
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wowbagger
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Re: Marching Army (I)
« Reply #3 on: May 7th, 2003, 2:35am » |
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Frankly, I don't think the argument in my last post is a good one.
However, I stand by my answer. This time, please consider the fraction of time elapsed when the messenger turns around and how far the army has gotten by that time.
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towr
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Re: Marching Army (I)
« Reply #4 on: May 7th, 2003, 5:33am » |
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As I said, the speeds are relative to the movement of the army.
when he runs in the same direction as the army he has to run much more than a mile, since the army is moving in that same direction, relatively slowing him down in doing his first 'mile'.
Going back he gets an advantage since the army is moving away from him, but he benefits much less from it since the time they're alongside each other is much shorter. Therefor the answer must be more than 2 miles.
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wowbagger
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Re: Marching Army (I)
« Reply #5 on: May 7th, 2003, 6:12am » |
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on May 7th, 2003, 5:33am, towr wrote:| As I said, the speeds are relative to the movement of the army. |
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Sorry, for some obscure reason I didn't notice that part of your first post. 
Quote:| Therefor the answer must be more than 2 miles. |
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That's quite plausible. I don't feel like working it out on paper myself right now. 
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rloginunix
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Re: Marching Army (I)
« Reply #6 on: Jul 11th, 2015, 11:08am » |
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"working it out on paper" we recall the non-relativistic velocities addition rule:
the velocity of a point relative to the frame at rest is the vector sum of the velocity of the point relative to the frame in motion and the velocity of the frame in motion relative to the frame at rest:
Let the ground (YOX) be the frame at rest, messenger - the point, the army - the frame in motion. Initially both frames are at O (and O'). After some time t the frame in motion is at O' (upper left drawing) and:
V(A, OX) is the velocity of the frame in motion relative to the frame at rest (Army/ground);
u(M, OX) is the velocity of the point relative to the frame at rest (Messenger/ground);
v(M, A) is the velocity of the point relative to the frame in motion (Messenger/Army):
u = v + V (in vector form)
Projecting the vectors on OX we get:
u = v + V
v = u - V
After the first rendezvous at R1 - when the messenger runs towards [/b]R2[/b] - we have the drawing on the upper right. Projecting its vectors on OX we get:
-u = -v + V
v = u + V
A frame is a coordinate system plus a clock so an observer in the frame in motion (at army's head) will clock the (first) arrival of the messenger at L/v = L/(u - V), an observer at army's rear will clock the (second) arrival of the messenger at L/v = L/(u + V) and an observer in the frame at rest at F (lower drawing) will clock the arrival of the army's head at L/V, where L is the length of the army's formation (in either frame) and u = "m", V = "a" in towr's notation.
Since the speeds here are << c the clocks tick the same in all the frames, hence - towr's equation.
It is also possible to solve this problem in the frame at rest. In it, at R1 the army covers x units of space, the messenger covers (L + x), the observer at R1 clocks both in at:
(L + x)/u = x/V (1)
By the time the messenger arrives at R2, covering x, the army's head arrives at F, rear - at R2, covering the remaining (L - x) units of space (all - in the frame at rest):
x/u = (L - x)/V (2)
Solve (1) and (2) for x and equate them to find u:
x = L*V/(u - V) = L*u/(u + V) = x
u2 - 2uV + Vu2 = 0
u = V(1 +  2)
x = L (2)/2
The distance d(M, OX) covered by the messenger relative to the ground then is:
L + 2x = L(1 + 2)
Seems like a lot of work for a simple problem but the formality of a method knows nothing about the problem it is applied to and things may get interesting fast even with the slightest of rearrangements.
A vector is a magnitude plus direction so let us keep everything the same but change only one parameter for only one object: the army still travels along OX with V - constant magnitude and direction (known), the messenger travels with u - constant magnitude only (given). Initially - army's head is at O, messenger - is on OY at Y0 above O - known. Then, as army moves along OX at constant velocity V, the vector of the messenger's velocity always points at the army's head while its magnitude remains the same. u > V. Find: the rendezvous time, the y(x) curve traced by the messenger.
(I was searching the forum for a potential duplicate of an army-related problem when I came across this one so I decided to add a different solution. Please look for "10,000 - 5,000" in easy)
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