What is the start time of the snowstorm?

[Ackbach]

Here is this week's POTW:

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It has been snowing heavily and steadily. At noon, a snow plow starts out and travels 2 miles the first hour, and 1 mile the second hour. When did it start snowing?

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[Ackbach]

Congratulations to MarkFL for his correct solution, which follows:

Spoiler

We should be able to reasonably assume the snow plow can clear a constant $P$ volume of snow per time, which we will measure in cubic feet per minute. If we let $x(t)$ be the displacement of the plow at time $t$ (in ft/min), $w$ be the width of the plow's blade, and $D(t)$ be the depth of the snow at time $t$ (in ft) then we may state:

\(\displaystyle wD(t)\d{x}{t}=P\tag{1}\)

Let $r$ be the rate at which the snow is falling, in ft/min. Let $t=0$ be noon and $D_0=D(0)$, and so we may write:

\(\displaystyle D(t)=rt+D_0\)

And so we may arrange (1) as:

\(\displaystyle \d{x}{t}=\frac{P}{w\left(rt+D_0\right)}\)

Integrating with respect to $t$, we obtain:

\(\displaystyle x(t)=\frac{P}{wr}\left(\ln\left|rt+D_0\right|+C\right)\)

Assuming $x(0)=0$, we then find:

\(\displaystyle 0=\frac{P}{wr}\left(\ln\left(D_0\right)+C\right)\implies C=-ln\left(D_0\right)\)

And so we now have:

\(\displaystyle x(t)=\frac{P}{wr}\ln\left(\frac{r}{D_0}t+1\right)\)

Let \(\displaystyle k_1=\frac{P}{wr}\) and \(\displaystyle k_2=\frac{r}{D_0}\)

and we may state:

\(\displaystyle x(t)=k_1\ln\left(k_2t+1\right)\)

We are told $x(60)=10560$ and $x(120)=15840$ and so this gives rise to the system:

\(\displaystyle k_1\ln\left(60k_2+1\right)=10560\)

\(\displaystyle k_1\ln\left(120k_2+1\right)=15840\)

The first equation implies:

\(\displaystyle k_1=\frac{10560}{\ln\left(60k_2+1\right)}\)

And substituting into the second, we obtain:

\(\displaystyle \frac{10560}{\ln\left(60k_2+1\right)}\ln\left(120k_2+1\right)=15840\)

\(\displaystyle 2\ln\left(120k_2+1\right)=3\ln\left(60k_2+1\right)\)

\(\displaystyle \left(120k_2+1\right)^2=\left(60k_2+1\right)^3\)

Discarding all but the positive roots, we find:

\(\displaystyle k_2=\frac{r}{D_0}=\frac{1+\sqrt{5}}{120}\)

Now setting $D(t)=0$, we find:

\(\displaystyle rt+D_0=0\)

\(\displaystyle \frac{r}{D_0}t=-1\)

Substituting in the value we found above, we have:

\(\displaystyle \frac{1+\sqrt{5}}{120}t=-1\)

And solving for $t$, we obtain:

\(\displaystyle t=-\frac{120}{1+\sqrt{5}}=30(1-\sqrt{5})\approx-37.082039325\)

Thus, we may conclude that it began snowing at (to the nearest second):

\(\displaystyle 11:22:55\text{ am}\)