This article is from the Puzzles FAQ, by Chris Cole email@example.com and Matthew Daly firstname.lastname@example.org with numerous contributions by others.
Snow starts falling before noon on a cold December day. At noon a
snow plow starts plowing a street. It travels 1 mile in the first hour,
and 1/2 mile in the second hour. What time did the snow start
You may assume that the plow's rate of travel is inversely proportional
to the height of the snow, and that the snow falls at a uniform rate.
Let b = the depth of the snow at noon, a = the rate of increase in the
depth. Then the depth at time t (where noon is t=0) is at+b, the
snowfall started at t_0=-b/a, and the snowplow's rate of progress is
ds/dt = k/(at+b).
If the snowplow starts at s=0 then s(t) = (k/a) log(1+at/b). Note that
s(2 hours) = 1.5 s(1 hour), or log(1+2A/b) = 1.5 log(1+A/b), where
A = (1 hour)*a. Letting x = A/b we have (1+2x)^2 = (1+x)^3. Solve for
x and t_0 = -(1 hour)/x.
The exact answer is 11:(90-30 Sqrt).
_American Mathematics Monthly_, April 1937, page 245
E 275. Proposed by J. A. Benner, Lafayette College, Easton. Pa.
The solution appears, appropriately, in the December 1937 issue,
pp. 666-667. Also solved by William Douglas, C. E. Springer,
E. P. Starke, W. J. Taylor, and the proposer.
See R.P. Agnew, "Differential Equations," 2nd edition, p. 39 ff.