The Snowplow Problem: When Did It Start Snowing?
One morning it starts snowing at a heavy and constant rate. At noon a snowplow sets out to clear a road. The plow removes snow at a constant rate: the volume of snow it clears per unit time is the same at all times, so the deeper the snow, the slower the plow moves.
By 1:00 pm the plow has traveled $2$ miles; by 2:00 pm it has traveled a total of $3$ miles.
At what time did it start snowing?
Hints
- Let the snow have started $T$ hours before noon, and measure time $t$ in hours after noon. Since it snows at a constant rate, the depth at time $t$ is proportional to $t + T$.
- Constant volume cleared per hour means (speed) times (depth) is constant, so the plow's speed is $v(t) = \dfrac{k}{t + T}$. Distance traveled is the integral of $v$.
- Write $\int_0^1 \frac{k\,dt}{t+T} = 2$ and $\int_0^2 \frac{k\,dt}{t+T} = 3$. Divide to eliminate $k$, then clear the logarithms by exponentiating: $\left(\frac{2+T}{T}\right)^2 = \left(\frac{1+T}{T}\right)^3$.
Worked Solution
How to Think About It: Translate the two physical statements into an ODE. Constant snowfall makes depth linear in elapsed time; constant clearing rate makes speed inversely proportional to depth. Integrating the speed gives distance as a logarithm, and the two observed distances leave one unknown, the start time.
Quick Estimate: In the first hour the plow covers $2$ miles, in the second only $1$ mile. If snow had started at noon exactly the speed would be infinite at the start (depth zero), and if it had started many hours before noon the depth would barely change between $12$ and $2$ and the two hourly distances would be nearly equal. The observed $2:1$ ratio sits between those extremes, suggesting the snow started well under an hour before noon.
Formal Solution:
*Step 1 -- Model.* Let snow begin $T$ hours before noon and let $t \ge 0$ be hours after noon. Snow depth is $h(t) = c\,(t + T)$ for a constant $c$. The plow clears a constant volume per hour, so $v(t)\,h(t) = \text{const}$, giving $$v(t) = \frac{k}{t + T}$$ for some constant $k > 0$.
*Step 2 -- Distance traveled.* $$D(t) = \int_0^{t} \frac{k\,ds}{s + T} = k \ln\frac{t + T}{T}.$$
*Step 3 -- Impose the observations.* $$k \ln\frac{1 + T}{T} = 2, \qquad k \ln\frac{2 + T}{T} = 3.$$ Dividing the second by the first eliminates $k$: $$\frac{\ln\frac{2+T}{T}}{\ln\frac{1+T}{T}} = \frac{3}{2} \quad \Longrightarrow \quad \left(\frac{2 + T}{T}\right)^{2} = \left(\frac{1 + T}{T}\right)^{3}.$$
*Step 4 -- Solve.* Multiply through by $T^3$: $$T(2 + T)^2 = (1 + T)^3 \;\Longrightarrow\; T^3 + 4T^2 + 4T = T^3 + 3T^2 + 3T + 1 \;\Longrightarrow\; T^2 + T - 1 = 0.$$ The positive root is $$T = \frac{\sqrt{5} - 1}{2} \approx 0.618 \text{ hours} \approx 37.1 \text{ minutes}.$$
*Step 5 -- Interpret.* The snow started about $37$ minutes before noon, i.e. at roughly $11{:}23$ am. (For completeness, $k = 2/\ln\frac{1+T}{T} = 2/\ln(1 + \phi) \approx 2.08$ miles, where $\phi = (1+\sqrt5)/2$ and $\frac{1+T}{T} = \phi^{2}$.)
Answer: Snow started $T = \dfrac{\sqrt{5} - 1}{2}$ hours $\approx 37$ minutes before noon, at about $11{:}23$ am.
Intuition
The plow's speed is inversely proportional to how long it has been snowing, so distance grows like a logarithm of elapsed time, and the two distance readings pin down the unknown start through a ratio of logs. Modelling a rate that is inversely proportional to an accumulating stock is the same structure as market impact models where execution speed falls as accumulated inventory grows, and the golden-ratio answer is a nice reminder that clean numbers hide in messy word problems.