redsturgeon wrote:This is the problem to be solved:
"One day it started snowing in the morning at a heavy and steady rate. A snowplough started out at noon, going two miles in the first hour and one mile in the second hour. What time did it start snowing?"
This is not a trick question.
Here is my solution, on UncleEbenezer's assumption that the snowplough clears snow at a fixed rate, travelling at the speed needed to encounter snow at that rate (and also UncleEbenezer's other assumptions, but those are standard 'geometrical perfection' assumptions for this sort of problem):
Define a unit of snow to be the amount of snow falling per hour and per mile of road, and define our time system so that 'time 0' is when the snow starts falling and 'time 1' is one hour later. We don't know what time noon is in that system, so define time N to be noon. Then the snowplough has travelled the first two miles by time N+1 and the next mile by time N+2. We know that the snow started falling in the morning, so N is positive (a good thing too, because if our hypothetical snowplough started moving before the snow started falling, it would set off at an infinite speed!).
Let the fixed rate at which the snowplough is clearing snow be F units per hour. At time t, the road ahead of the snowplough has accumulated a snow load of t units per mile, and so the snowplough is making progress through it at F/t miles per hour. Between times t0 and t1, it will travel a distance equal to the integral from t=t0 to t=t1 of that speed F/t. The indefinite integral of 1/t is ln(t), so the indefinite integral of F/t is F*ln(t), and the distance travelled is that indefinite integral evaluated at t=t1 minus that indefinite integral evaluated at t=t0, i.e. F*ln(t1) - F*ln(t0) = F * (ln(t1)-ln(t0)).
So the distance travelled in the first hour is F * (ln(N+1)-ln(N)), and the distance travelled in the second hour is F * (ln(N+2)-ln(N+1)). But we're told that the first of those is 2 and the second 1, so:
F * (ln(N+1)-ln(N)) = 2 * F * (ln(N+2)-ln(N+1))
Dividing both sides by F and rearranging terms, this simplifies to:
3 * ln(N+1) = ln(N) + 2 * ln(N+2)
Taking e to the power of both sides gives:
(N+1)^3 = N * (N+2)^2
or after multiplying that out:
N^3 + 3N^2 + 3N + 1 = N^3 + 4N^2 + 4N
Subtracting N^3 + 3N^2 + 3N from both sides leaves:
1 = N^2 + N
a quadratic equation with the two solutions (-1 +/- sqrt(5))/2 by the standard formula. The solution (-1 - sqrt(5))/2 is negative, which N cannot be, so N = (sqrt(5) - 1)/2 = 0.6180339887... So the answer is that it started snowing that number of hours before noon, or to the nearest second, 37 minutes and 5 seconds before noon, i.e. at 11:22:55 am.
Having watched that after writing out my solution, it's essentially the same solution as mine, but with some minor transformations - e.g. it uses a time origin of noon, I use the time the snow starts falling; it chooses a slightly different way of presenting the use of standard results about logarithms to me; I take an unknown constants out of the formulae by defining a unit of snow to make one of those constants equal to 1.
Gengulphus