# That's my name, Mr. Plow

by araya

At time t, the amount of snow is some constant(r) times t.
snow = rt
The velocity at which the snowplow can travel is inversely proportional to the amount of snow:
v = c/rt
Since the plow travelled the same distance from noon to 1 as it did from 1 to 4, we can set these distances equal, and use the integral of the velocity as the distance. The snow started at some time x before the plow, so the integrals are from x to 1+x, and from 1+x to 4+x. These integrate to:
r/c(ln(1+x) - ln(x)) = r/c(ln(4+x) - ln(1+x))
The constants cancel, and rearranging use logarithmic properties gives:
ln((4+x)(x)/(1+x)^2) = 0
(4+x)(x)/(1+x)^2 = 1
(4+x)(x) = (1+x)^2
Which simplifies to x=1/2. So, the snow started at 11:30 AM.