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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 t1 seconds after it starts snowing, and the second starts from the same point t2 t1 seconds later, going in the same direction. Snow falls so that the depth of snow increases at a constant rate of k ms1. The speed of each snowplough is akz ms1 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 dt dx = t t1exa. ()

Hence show that the snowploughs will collide when they have travelled a(t2t1 1) 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 () can be solved using an integrating factor. However, the equation which arises naturally at this point is one involving dx dt , 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 .

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(kT), i.e. aT. The equation of motion of the first snowplough is

dx dT = a T .

Integrating both sides with respect to T gives

x = alnT + constant of integration.

We know that T = t1 when x = 0 (the snowplough started t1 seconds after the snow started), so

x = alnT alnt1.

This can be rewritten as

T = t1exa.

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 t1exa) metres. Thus the equation of motion of the second snowplough is

dx dt = ak k(t t1exa).

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

dx dt = 1 dt dx

to obtain the required equation ().

Multiplying by exa (an integrating factor) and rearranging gives

exa dt dx exat a = t1 a      i.e.       d dx texa = t1 a

which integrates to

texa = t1 a x + constant of integration.

Since the second snowplough started (x = 0) at time t2, the constant of integration is just t2 and the solution is

t = (t2 t1xa)exa.

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

T = tt1eXa = (t 2 t1Xa)eXa,

so X is given by

t1 = (t2 t1Xa).

This is equivalent to the given formula.