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One day snow began to fall before dawn and continued to fall at a constant rate. At midday a snowplot set out to clear a road. At 2pm it turned back, arriving to the starting point at 3pm. If we suppose that the snowplow speed is inversely proportional to the height of the snow, at what time did it start snowing? At what time should the snowplot turn back in order to arrive to the starting point at 2pm?


$x(t)$ be the position of the snowplow at time $t$.

$h(t)$ the height of the snow at time $t$

We denote $\Delta V $ the volume of snow removed by the snowplow in a time $\Delta t$ small enough to suppose that $h(t)$ constant in $\Delta x$. If L is the widht of the shovel of the snowplow we have:

$\frac{\Delta V}{\Delta t} = h L \frac{\Delta x}{\Delta t}$

If we take $\Delta t \rightarrow 0$:

$\frac{dV}{dt} = h L \frac{dx}{dt}$

We know that $\frac{dV}{dt} = \alpha$ is constant, so we have

$\frac{dx}{dt} = \frac{\alpha}{L h}$

As snow falls at a constant rate, we have $h(t) = c t$ then:

$\frac{dx}{dt} = \frac{\alpha}{L c t}$

Calling $ A = \frac{\alpha}{L c}$ we have:

$\frac{dx}{dt} = \frac{A}{t}$

and then:

$x(t) = A Log(t) + C$

We have $x(T) = 0 \Longrightarrow x(t) = A Log(\frac{t}{T})$

This is clearly the position of the snowplow between 12pm and 2pm. I don't know how to continue from here because in the way back h(t) isn't as simple and depends on when the snowplow went over there the first time.

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Is this homework? Please show us how you've set up the problem and how you're working on it. – Ted Shifrin May 30 '13 at 14:23
Well done on the set-up, although I might have said h(t) is assumed constant in the small time interval $\Delta t$. When does your clock start? Are you sure $h(t)=ct$? Next question, which I suspect may have puzzled you too: When the snowplow turns around, does he go over the same part of the road, or does he do the other side of the road? – Ted Shifrin May 30 '13 at 15:34
he goes over the same part of the road, that's the hard part. My clock start when it start snowing. – José D. May 30 '13 at 15:35
Ah, ok. So your formula for $x(t)$ holds for $T\le x\le T+2$. If he goes over the same part and we ignore the time it takes him to turn around, we do have a big problem, as at precisely 2:00 he is moving with infinite speed. On the other hand, if he travels on the other side, it will obviously take much longer to go that same distance. So how do we interpret this to have a real problem? – Ted Shifrin May 30 '13 at 15:44
If you assume that the snowplow turns back instantanely we do not have any trouble with that. I think it is quite obvious that it will take less time to go back. – José D. May 30 '13 at 16:26

The only sense I can make of the problem is to assume that the plow covers exactly the same route, so the snow depth between $2$ and $3$ is only the snow that fell since the plow passed over the point the time before. The logic for this comes from the fact that the plow made better time between $2$ and $3$ than it did between $12$ and $2$, so there must have been less snow on the ground. A rough approach is that there was an average of $1\frac 12$ hours of snow on the second pass, so there must have been an average of $3$ hours on the first pass, so it started snowing at $10$ AM. The alternate, that the plow turns around and comes back over the same route, fails because the speed would be infinite at the turnaround and the singularity is not integrable.

It would be good to define your variables explicitly. You seem to define $t=0$ as when it starts snowing, then $t=T$ as noon when the plow sets out. You are fine with $x(t)=A \log \frac tT$ between $12$ and $2$. Then between $2$ and $3$ you have the height is $c$ times the time between the first pass and the second pass. This should allow you to integrate it.

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I honestly don't follow. There is not a linear relation between depth and time to clear. The only thing I can think to do is to assume the plow can only clear up to a certain depth, and there is some snow still left on the ground as he turns around at 2 pm. I don't think there's enough information to solve this. – Ted Shifrin May 30 '13 at 17:35
@TedShifrin: You are right that there is not a linear relationship, but I was making an approximation. They often help in scaling the problem. My second paragraph indicates how to get the correct answer. We'll see how close it comes. I indicate why I don't think turning around makes a reasonable problem. – Ross Millikan May 30 '13 at 17:44

At first,I think your fomula $x(t)=Log( \frac{t}{T})$ is wrong , because there was snow in the road even before the snowplow works.

Denote $x(t),h(t)$ as you mentioned before.
Suppose$x(0)=0$; $t=0$ at $12pm$ and $t=t_1$ at $2 pm$.

Incase $t \le t_1$

Because the snow has alreaday been falling, we have $ h(t)=h_0+\lambda.t \forall t \in [0;t_1]$
Due to hypothesis , we have
$ \frac{dx}{dt}=\frac{C}{h(t)}= \frac{C}{h_0+\lambda.t} =\frac{C}{\lambda}.\frac{1}{\frac{h_0}{\lambda}+t}$ Therefore $ x(t)= \frac{C}{\lambda}.ln(1+\frac{\lambda}{h_0}.t)$.
In conclusion,
We have proved that the snowplow swept through point $x=x_0$ at the the time $t=\frac{h_0}{\lambda}.[ exp(x(t).\frac{\lambda}{C})-1]$ in case $t<t_1$

.In the case $t \in (t_1,+\infty)$:

When $t>t_1$ , $h(t)=\lambda.\{t-\frac{h_0}{\lambda}.[ exp(x(t).\frac{\lambda}{C})-1] \}$
($h$ is discontinuous at $t_1$,$x(t)$ is decreasing on $(t_1;\infty)$).
$\Leftrightarrow h_0+\lambda.t-h_0.exp(x.\frac{\lambda}{C})= -\frac{dt}{dx}$
I can only go thus far, and then consider $t$ as a function $t(x)$ in this case and use caculator to do the rest.

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Old problem, but I like the snowplow problem. The twist this time is that it replows the same lane on the return trip, else it would take longer than the trip out. It snows at constant rate: $$\frac{dh}{dt}=r$$ where $h$ is the snow depth, $t$ is the time, conventionally $t=0$ at midnight, and $r$ the the rate of snowfall. Then $h=rt+C_1$; at $t=t_0$ when it starts snowing, $0=rt_0+C_1$, so $h=r(t-t_0)$. Then if $x$ is its distance from its starting point, its speed is $$\frac{dx}{dt}=\frac kh=\frac k{t(t-t_0)}$$ where $k$ is the constant of proportionality. Separating variables and integrating, $$dx=\frac kr\frac{dt}{(t-t_0)}$$ $$x=\frac kr\ln(t-t_0)+C_2$$ At $t=t_1$ when it starts out, $$0=\frac kr\ln(t_1-t_0)+C_2$$ so $$x=\frac kr\ln\left(\frac{t-t_0}{t_1-t_0}\right)$$ We can solve this to find the time $\tau(x)$ when it passed point $x$: $$\tau(x)=t_0+(t_1-t_0)e^{\frac{rx}k}$$ On its return trip its path has only been snowed on since $\tau(x)$ so now $$h=r\left(t-t_0-(t_1-t_0)e^{\frac{rx}k}\right)$$ Also its distance to home is decreasing now, so $$\frac{dx}{dt}=\frac{-k}h=\frac{-k}{r\left(t-t_0-(t_1-t_0)e^{\frac{rx}k}\right)}$$ Here we diverge from @Toan Nguyen Dinh's solution because that author forgot about the constant of proportionality in the numerator. $$\frac{dt}{dx}+\frac rkt=\frac rk\left(t_0+(t_1-t_0)e^{\frac{rx}k}\right)$$ Applying the integrating factor $$\mu=e^{\int\frac rkdx}=e^{\frac{rx}k}$$ $$\frac d{dx}\left(e^{\frac{rx}k}t\right)=e^{\frac{rx}k}\frac{dt}{dx}+\frac rke^{\frac{rx}k}t=e^{\frac{rx}k}\frac rk\left(t_0+(t_1-t_0)e^{\frac{rx}k}\right)$$ Integrating, $$e^{\frac{rx}k}t=t_0e^{\frac{rx}k}+\frac12(t_1-t_0)e^{\frac{2rx}k}+C_3$$ At $t=t_2$ when it turns around, $$e^{\frac{rx}k}=\frac{t_2-t_0}{t_1-t_0}$$ $$\frac{(t_2-t_0)}{(t_1-t_0)}t_2=t_0\frac{(t_2-t_0)}{(t_1-t_0)}+\frac12(t_1-t_0)\left(\frac{t_2-t_0}{t_1-t_0}\right)^2+C_3$$ $$C_3=\frac12\frac{(t_2-t_0)^2}{(t_1-t_0)}$$ So when it returns to base at $t=t_3$, $x=0$, $$t_3=t_0+\frac12(t_1-t_0)+\frac12\frac{(t_2-t_0)^2}{(t_1-t_0)}$$ Simplifying, $$t_1t_3-\frac12t_2^2-\frac12t_2^2=(t_3-t_2)t_0$$ Since $t_1=12$, $t_2=14$, and $t_3=15$, $t_0=10$, so it started out at 10:00 AM. So the intuition of @Ross Millikan was good! Then if we wanted to get back at 2:00 PM, $t_3=14$, and $$\frac12t_2^2-t_0t_2+t_0t_3-t_1t_3+\frac12t_1^2=0$$ $$\frac12t_2^2-10t_2+44=0$$ $$t_2=\frac{10\pm\sqrt{100-88}}1=10+2\sqrt3$$ Because it didn't start plowing until after it began snowing. Converting to h:m:s, it had to turn around at 1:27:50.77 PM.

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