When did the snow start?

We’ll assume the snow starts at $t=0$ hours. So after $t$ hours, the depth of snow on the ground is proportional to $t$. Since the plow’s speed $v$ is inversely proportional to the amount of snow on the ground, we have $v = \tfrac{c}{t}$, where $c>0$ is a constant of proportionality. We’ll assume the constant is in units of miles so that $v$ is in miles per hour.

Suppose the plow starts at $t=x$ hours. During the first hour, the plow travels 2 miles. Since distance is the integral of velocity with respect to time, we have:
\[
2\text{ miles} = \int_{x}^{x+1} v\,\mathrm{d}t = \int_x^{x+1} \frac{c}{t}\,\mathrm{d}t = c\,\log\left( \frac{x+1}{x} \right)
\]During the second hour, the plow travels 1 mile. So we also have:
\[
1\text{ mile} = \int_{x+1}^{x+2} v\,\mathrm{d}t = \int_{x+1}^{x+1} \frac{c}{t}\,\mathrm{d}t = c\,\log\left(\frac{x+2}{x+1} \right)
\]Dividing the first equation by the second, the units and constants of proportionality cancel. Simplifying, we obtain:
\[
2\,\log\left(\frac{x+2}{x+1}\right) = \log\left( \frac{x+1}{x} \right)
\]Exponentiating both sides to cancel out the logs,
\[
\left(\frac{x+2}{x+1}\right)^2 = \frac{x+1}{x}
\]Finally, clearing fractions and simplifying, we obtain:
\[
x^2+x-1 = 0
\]Since we must have $x\gt 0$, we can discard the negative solution to this equation, and keep only the positive one. This leads us to:
\[
x = \frac{\sqrt{5}-1}{2} \approx 0.618034
\]Incidentally, this is the inverse of the Golden Ratio, though that’s more of a coincidence than anything meaningful… If the plow started at noon, then the snow started $x$ hours before noon, which is 11:22.55 AM (11 hours, 22 min, 55 sec).