<pre>
11:22:55.077 am.

Method:

Let b <code> the depth of the snow at noon, a </code> 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<code>0 then s(t) </code> (k/a) log(1+at/b).  Note that
s(2 hours) <code> 1.5 s(1 hour), or  log(1+2A/b) </code> 1.5 log(1+A/b), where
A <code> (1 hour)*a.  Letting x </code> 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(5)).

"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.
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