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Topic: Three Snow Ploughs (Read 800 times) |
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ThudnBlunder
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Three Snow Ploughs
« on: Feb 19th, 2009, 6:23am » |
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It began snowing at a constant rate before midday. Three ploughs set out at midday, 1.00pm, and 2.00pm, repectively, removing snow at a constant rate. If at some later time they all came together simultaneously, what was this time and when did it begin snowing?
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| « Last Edit: Feb 19th, 2009, 6:24am by ThudnBlunder » |
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balakrishnan
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Re: Three Snow Ploughs
« Reply #1 on: Feb 24th, 2009, 2:26pm » |
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Since no one has replied(including EigenRay), it clearly means that no one understood what "came together simultaneously" means.
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towr
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Re: Three Snow Ploughs
« Reply #2 on: Feb 24th, 2009, 2:29pm » |
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Yeah, I had been meaning to ask, are they driving the same stretch of road?
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Grimbal
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Re: Three Snow Ploughs
« Reply #3 on: Feb 24th, 2009, 2:44pm » |
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I guess that they follow the same stretch, #1 will slow down more and more and eventually #2, who benefits from the path opened by #1, will catch up on him. It seems under some circumstances, #3 will also catch up on #2. And in one particular case #3 will catch up on #2 exactly when #2 catches up on #1. You need to find in which case.
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Obob
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Re: Three Snow Ploughs
« Reply #4 on: Feb 24th, 2009, 7:39pm » |
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Although mustn't the velocity of the trailing plows tend to infinity in the moments before they catch up?
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Grimbal
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Re: Three Snow Ploughs
« Reply #5 on: Feb 25th, 2009, 2:01am » |
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Even twice infinity for the second one 
I guess you can solve it ignoring special relativity. Ignore also the physical size of the snow ploughs.
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ThudnBlunder
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Re: Three Snow Ploughs
« Reply #6 on: Mar 3rd, 2009, 11:35am » |
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on Feb 24th, 2009, 2:29pm, towr wrote:| Yeah, I had been meaning to ask, are they driving the same stretch of road? |
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Had more than one (stretch of) road been necessary, I feel sure the question would have mentioned them.
As it didn't, one can assume that one (narrow) road is sufficient.
on Feb 24th, 2009, 2:26pm, balakrishnan wrote:| Since no one has replied(including EigenRay), it clearly means that no one understood what "came together simultaneously" means. |
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It means, when considered as points, their position vectors coincide.
on Feb 25th, 2009, 2:01am, Grimbal wrote:| I guess you can solve it ignoring special relativity. |
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If this had been a challenge from a Bernoulli, I'm sure Isaac would have solved it in a jiffy. 
on Feb 24th, 2009, 7:39pm, Obob wrote:| Although mustn't the velocity of the trailing plows tend to infinity in the moments before they catch up? |
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They meet before infinity becomes a non-issue.
Consider dx/dt, dy/dt, dz/dt, where
x,y,z are respective distances travelled in t hours past midday.
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| « Last Edit: Mar 17th, 2009, 3:08pm by ThudnBlunder » |
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balakrishnan
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Re: Three Snow Ploughs
« Reply #7 on: Mar 4th, 2009, 7:26am » |
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It started snowing at 11:20 am
and they met at 2:19:16 pm (ie 2hours, 19 minutes and 16 seconds)
Let s be the rate at which it snows. Let R be the rate at which the plough removes the snow.
Consider the first plough,
Let the position be p1
We see that the equation of motion would be
dp1/dt= R/[s(t+t0)]
where t is the time (wrt 12 noon) and 12-t0 is the time at which it started snowing.
Solving gives
t1=t0(esp1/R-1)
For the second plough ,
let p2 be the position.
We see that dp2/dt is R/[s(t-t1(p2))]
With boundary conditions p2(1)=0, we get
t2=(1+t0-st0 p2/R)esp2/R-t0
Similarly for the third plough
t3=-t0+esp3/R [0.5(st0 p3/R)2+2+t0-s(1+t0)p3/R)
Solving t1=t2=t3
gives t0=2/3
or it started snowing at 11:20 am
and they met somewhere between 2:19:15:53 pm and 2:19:15:54 pm(ie 2hours, 19 minutes,15 seconds and 53 ms)
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| « Last Edit: Mar 4th, 2009, 8:22am by balakrishnan » |
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ThudnBlunder
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balakrishnan, I think I have the same equations as you, but not exactly the same conclusion.
Let s/R = k
Let n be the number of hours before midday that it starts snowing.
Let x,y,z be the respective distances travelled in t hours past midday.
Then dt/dx = k(t + n), with x = 0 when t = 0
Giving t = n(ekx - 1)
And dt/dy = t - n(eky - 1), with y = 0 when t = 1
Giving t = eky[1 + n - ny] - n
And dt/dz = t - ez[n(1 - z) + 1] - n, with z = 0 when t = 2
Giving t = ekz[0.5nz2 - (n + 1)z + n + 2] - n
Let d = the common value of x,y,z when the ploughs meet.
Let T = the value of t when the ploughs meet.
Then (T + n)/ekd = n = 1 + n - nd = 0.5nd2 - (n + 1)d + n + 2
Leading to
n = 1/2
d = 2
T = (e2k - 1)/2
So the time of meeting would seem to depend on the ratio s/R = k
Putting k = 1 gives
T = (e2 - 1)/2
= 3.194528...
In the graph below k = 1 and n = 1/2
CONCLUSION: The 2nd plough was manufactured in a Communist country, after all. LOL
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| « Last Edit: May 28th, 2011, 6:37am by ThudnBlunder » |
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