Snow-plough puzzle, difficult variation

I checked the archive. This puzzle does not appear anywhere as complicated as this.

1. Snow falls at constant rate throughout our story.

2. Speed of a Canadian snow-plough is inversely proportional to the instantaneous depth of snow being ploughed.

3. At 1 a.m. a plough starts out and travels in a straight line.

4. At 2 a.m. a second plough starts out from same base, following in the tracks of the first.

5. And at 3 a.m. a third plough does likewise, in the tracks of the second.

6. Eventually, all three ploughs collide together.

At what time did it start to snow, and at what time did the crash take place?

Is this too hard for this NG?

Cheeky

I don't know about this. The second plough starts off ploughing 1 hour's worth of snow, and goes on its merry way. An hour later, the third plough also starts off ploughing 1 hour's worth of snow, so its speed as a function of the time since it started should be identical to that of the second plough. So it looks to me like the second plough maintains a constant 1-hour lead on the third plough (though the *distance* between them will shrink, but never to zero).

In math:

Let a be the time (in hours past midnight) is started snowing.

Then v1(t) = 1/(t-a) for t >= 1, v2(t) = 1/(t-1) for t >= 2, v3(t) = 1/(t-2) for t >= 3,

since plough 1 starts off with 1-a hours of snow, plough 2 with 2-1 hours of snow, and plough 3 with 3-2 hours of snow.

So, integrating, x1(t) = ln(t-a) - ln(1-a) for t >= 1, x2(t) = ln(t-1) for t >= 2, x3(t) = ln(t-2) for t >= 3,

since x1(1) = x2(2) = x3(3) = 0.

So it looks to me like ploughs 2 and 3 will never collide.

- Ian

Perhaps the top speed of each of the plows is different.

The second and third ploughs speed function won't be the same. Each ploughs function is related to how fast the plough in front has travelled because the snow fallen for the next plough is only the part that falls <behind> the one in front...

Can I borrow one please? There is hardly ever settled snow here, and it would be pretty speedy for getting around...

Right. One hour's worth, in each case.

- Ian

OK, good point. In that case, let the relative top speeds be 1,b,c.

Then ln(t-a) - ln(1-a) = b ln(t-1) = c ln(t-2)

I don't think there's enough information to solve for 4 variables with justr the 2 equations.

If the third plough started at, say, 2:30, the problem would be solvable.

- Ian

Yes, but the hours worth has fallen in different amounts at each distance because the one in front has travelled at a different speed. Since the speed is <inversely> proportional, adding up the speeds etc will not give you the same totals.

I can tell you that the problem is solveable, according to my very reliable PhD source. I gather an undergraduate can follow the solution, but probably not come up with it himself Good luck Stan

but On Thu, 22 Nov 2001 22:24:54 +0000 (UTC),

It was while I was trying to compose a post to show you the error of your logic, I finally saw the error of mine.

In my book plough 2 hits plough 1, but plough 3 is a long way behind (as plough 2 was speeding up, and leaving more and more distance thus more snow behind it).

I have a feeling that plough 4 would hit plough 3.

Question - do the helicopter recovery crew have to have removed ploughs 1 and 2 before plough 4 hits plough 3?