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        riddles >> medium >> Three Snow Ploughs
        
(Message started by: ThudanBlunder on Feb 19^th, 2009, 6:23am)

Title: Three Snow Ploughs
Post by ThudanBlunder on Feb 19^th, 2009, 6:23am

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?

Title: Re: Three Snow Ploughs
Post by balakrishnan on Feb 24^th, 2009, 2:26pm

Since no one has replied(including EigenRay), it clearly means that no one understood what "came together simultaneously" means.

Title: Re: Three Snow Ploughs
Post by towr on Feb 24^th, 2009, 2:29pm

Yeah, I had been meaning to ask, are they driving the same stretch of road?

Title: Re: Three Snow Ploughs
Post by Grimbal on Feb 24^th, 2009, 2:44pm

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.

Title: Re: Three Snow Ploughs
Post by Obob on Feb 24^th, 2009, 7:39pm

Although mustn't the velocity of the trailing plows tend to infinity in the moments before they catch up?

Title: Re: Three Snow Ploughs
Post by Grimbal on Feb 25^th, 2009, 2:01am

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.

Title: Re: Three Snow Ploughs
Post by ThudanBlunder on Mar 3^rd, 2009, 11:35am

on 02/24/09 at 14:29:06, towr wrote:

Yeah, I had been meaning to ask, are they driving the same stretch of road?

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 02/24/09 at 14:26:21, balakrishnan wrote:

Since no one has replied(including EigenRay), it clearly means that no one understood what "came together simultaneously" means.

It means, when considered as points, their position vectors coincide.

on 02/25/09 at 02:01:24, Grimbal wrote:

I guess you can solve it ignoring special relativity.

If this had been a challenge from a Bernoulli, I'm sure Isaac would have solved it in a jiffy. :P

on 02/24/09 at 19:39:29, Obob wrote:

Although mustn't the velocity of the trailing plows tend to infinity in the moments before they catch up?

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.

Title: Re: Three Snow Ploughs
Post by balakrishnan on Mar 4^th, 2009, 7:26am

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(e^sp₁/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)e^sp₂/R-t₀

Similarly for the third plough
t₃=-t₀+e^sp₃/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)

Title: Re: Three Snow Ploughs
Post by ThudanBlunder on Mar 17^th, 2009, 5:37pm

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(e^kx - 1)

And dt/dy = t - n(e^ky - 1), with y = 0 when t = 1
Giving t = e^ky[1 + n - ny] - n

And dt/dz = t - e^z[n(1 - z) + 1] - n, with z = 0 when t = 2
Giving t = e^kz[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)/e^kd = n = 1 + n - nd = 0.5nd² - (n + 1)d + n + 2
Leading to
n = 1/2
d = 2
T = (e^2k - 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