## Problem

One morning, it starts snowing. The snow falls at a constant rate, and it continues the rest of the day.

At noon, a snowplow begins to clear the road. The more snow there is on the ground, the slower the plow moves. In fact, the plow’s speed is inversely proportional to the depth of the snow — if you were to double the amount of snow on the ground, the plow would move half as fast.

In its first hour on the road, the plow travels 2 miles. In the second hour, the plow travels only 1 mile.

When did it start snowing?

## Solution

We have \[ \begin{align} \frac{dx}{dt} = \frac{k}{rt} \end{align} \] where \(k\) is a constant and \(r\) is the constant rate of snow fall.

Integrating the above equation we have the following equations for the distances covered in the first and second hours by the snowplow \[ \begin{align} 2 &= \int_{t_s}^{t_s+1} \frac{k}{rt} dt = \frac{k}{r}ln\left(\frac{t_s+1}{t_s}\right) \\ 1 &= \int_{t_s+1}^{t_s+2} \frac{k}{rt} dt = \frac{k}{r}ln\left(\frac{t_s+2}{t_s+1}\right) \end{align} \] where \(t_s\) is the time in hrs the snowplow starts after the snow fall begins.

From the above we have, \[
\begin{align}
\left(\frac{t_1+2}{t_s+1}\right)^2 = \frac{t_s+1}{t_s} \implies t_s = 0 \lor t_s^2+t_s-1 = 0
\end{align}
\] Ignoring the trivial and the implausible solutions, we have, \[
\begin{align}
t_s = \frac{-1 + \sqrt{5}}{2} = \frac{1}{\phi}
\end{align}
\] As the snowplow starts at noon, the snow fall must have started at \(12:00 - t_s\)hrs i.e. at *11:23 am*.