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Problem 52:  Snowploughing ($✓$ $✓$ $✓$) 1987 Specimen Paper III

Two identical snowploughs plough the same stretch of road. The ﬁrst starts at a time ${t}_{1}$ seconds after it starts snowing, and the second starts from the same point ${t}_{2}-{t}_{1}$ seconds later, going in the same direction. Snow falls so that the depth of snow increases at a constant rate of $k$ ms${}^{-1}$. The speed of each snowplough is $ak∕z$ ms${}^{-1}$ where $z$ is the depth (in metres) of the snow it is ploughing and $a$ is a constant. Each snowplough clears all the snow. Show that the time $t$ at which the second snowplough has travelled a distance $x$ metres satisﬁes the equation

 $a\frac{\mathrm{d}t}{\mathrm{d}x}=t-{t}_{1}{e}^{x∕a}.$ (†)

Hence show that the snowploughs will collide when they have travelled $a\left({t}_{2}∕{t}_{1}-1\right)$ metres.

There is something exceptionally beautiful about this question, but it is hard to identify exactly what it is; seeing the question for the ﬁrst time makes even hardened mathematicians smile with pleasure.

There is a modelling element to it: you have to set up equations from the information given in the text. The ﬁrst equation you need is a simple ﬁrst order differential equation to ﬁnd the time taken by the ﬁrst snowplough to travel a distance $x$. The corresponding equation for the second snowplough is a bit more complicated, because the depth of snow at any point depends on the time at which the ﬁrst snowplough reached that point, clearing the snow.

The differential equation $\left(†\right)$ can be solved using an integrating factor. However, the equation which arises naturally at this point is one involving $\frac{\mathrm{d}x}{\mathrm{d}t}\phantom{\rule{0.3em}{0ex}}$, which cannot (apparently) be solved by any means. It is the rather good trick of turning the equation upside down (regarding $t$ as a function of $x$ instead of $x$ as a function of $t$) that allows the problem to be solved so neatly. Apologies if you haven’t come across integrating factors for ﬁrst order differential equations; they are not on our syllabus, but they are really not difficult — you can look online and ﬁnd an easily understandable explanation.

You won’t surprised to learn that there is a generalisation to $n$ identical snowploughs $\dots$ .

Solution to problem 52

Suppose that the ﬁrst snowplough reaches a distance $x$ at time $T$ after it starts snowing. Then the depth of snow it encounters is $kT$ and its speed is therefore $ak∕\left(kT\right)$, i.e. $a∕T$. The equation of motion of the ﬁrst snowplough is

$\frac{\mathrm{d}x}{\mathrm{d}T}=\frac{a}{T}.$

Integrating both sides with respect to $T$ gives

We know that $T={t}_{1}$ when $x=0$ (the snowplough started ${t}_{1}$ seconds after the snow started), so

$x=alnT-aln{t}_{1}.$

This can be rewritten as

$T={t}_{1}{e}^{x∕a}\phantom{\rule{0.3em}{0ex}}.$

When the second snowplough reaches $x$ at time $t$, snow has been falling for a time $t-T$ since it was cleared by the ﬁrst snowplough, so the depth at time $t$ is $k\left(t-T\right)$ metres, i.e. $k\left(t-{t}_{1}{e}^{x∕a}\right)$ metres. Thus the equation of motion of the second snowplough is

$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{ak}{k\left(t-{t}_{1}{e}^{x∕a}\right)}\phantom{\rule{0.3em}{0ex}}.$

Now we use the standard result (a special case of the chain rule)

$\frac{\mathrm{d}x}{\mathrm{d}t}=1/\frac{\mathrm{d}t}{\mathrm{d}x}$

to obtain the required equation $\left(†\right)$.

Multiplying by ${e}^{-x∕a}$ (an integrating factor) and rearranging gives

which integrates to

Since the second snowplough started ($x=0$) at time ${t}_{2}$, the constant of integration is just ${t}_{2}$ and the solution is

$t=\left({t}_{2}-{t}_{1}x∕a\right){e}^{x∕a}.$

The snowploughs collide when they reach the same position at the same time. Let this position be $x=X$. Then

$T=t⇒{t}_{1}{e}^{X∕a}=\left({t}_{2}-{t}_{1}X∕a\right){e}^{X∕a},$

so $X$ is given by

${t}_{1}=\left({t}_{2}-{t}_{1}X∕a\right).$

This is equivalent to the given formula.