Problem 52

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 $a k ∕ z$ 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{d t}{d x} = t - t_{1} e^{x ∕ a} .

(†)

Hence show that the snowploughs will collide when they have travelled $a (t_{2} ∕ t_{1} - 1)$ metres.

Comments

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 $(†)$ can be solved using an integrating factor. However, the equation which arises naturally at this point is one involving $\frac{d x}{d t}$ , 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 $k T$ and its speed is therefore $a k ∕ (k T)$ , i.e. $a ∕ T$ . The equation of motion of the ﬁrst snowplough is

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

Integrating both sides with respect to $T$ gives

x = a ln T + constant of integration .

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

x = a ln T - a ln t_{1} .

This can be rewritten as

T = t_{1} e^{x ∕ a} .

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 $k (t - T)$ metres, i.e. $k (t - t_{1} e^{x ∕ a})$ metres. Thus the equation of motion of the second snowplough is

\frac{d x}{d t} = \frac{a k}{k (t - t_{1} e^{x ∕ a})} .

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

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

to obtain the required equation $(†)$ .

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

e^{- x ∕ a} \frac{d t}{d x} - \frac{e^{- x ∕ a} t}{a} = - \frac{t_{1}}{a} i.e. \frac{d}{d x} (t e^{- x ∕ a}) = - \frac{t_{1}}{a}

which integrates to

t e^{- x ∕ a} = - \frac{t_{1}}{a} x + constant of integration .

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

t = (t_{2} - t_{1} x ∕ a) 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 \Rightarrow t_{1} e^{X ∕ a} = (t_{2} - t_{1} X ∕ a) e^{X ∕ a},

so $X$ is given by

t_{1} = (t_{2} - t_{1} X ∕ a) .

This is equivalent to the given formula.