Question

Two Snowplows: One day it began to snow exactly at noon at a heavy and steady rate. A snow plow left its garage at 1:00 pm, and another followed in its tracks at 2:00 pm. (a) At what time did the second snowplow crash into the first? To answer this questions, assume as in Project D that the rate (in mph) at which a snowplow can clear the road is inversely proportional to the depth fo the snow (and hence to the time elapsed since the road was clear of snow). [Hint: Begin by writing dierential equations for x(t) and y(t), the distances traveled by the first and second snowplows, respectively, at t hours past noon. To solve the dierential equation involving y, let t rather than y be the dependent variable!]

(b) Could the crash have been avoided by dispatching the second snowplow at 3:00 pm instead?

please show all work

Answer 1

Let midday be t = 0 (in hours), and let the first and second snowplows cover distances x(t) and y(t) by time t. Then the first snowplow starts at t=1, and the second at time t=2.

The distance covered by the first snowplow follows an equation of the form

where h₁(t) is the height of snow in front of the snowplow at time t. It is clear that, since the snow started falling at midday,

For some constant rate of fall of snow r (in appropriate units). Hence

where γ = k/r. Integrating, and using the condition x(1) = 0, we get

After the first snowplow has cleared away the snow, the second snowplow faces a more complicated snow height profile h₂(y,t), depending on at what time the first snowplow had passed position y. Inverting (#), we see that

Hence the differential equation for the second snowplow is

This can be solved by the substitution z = y/γ :

where C is a constant. We could find C by noting that y(2) = 0, i.e. z(2) = 0; the result is C=2. Hence the solution is

Let d be the distance such that the time expressions for the two plows are the same numerically. Then setting z = d/γ, we get, comparing the (##) and the inverse of (#),

Hence the crash happened at 2.71828182846 hours after midday, i.e. 2 hours and 60*0.71828182846 = 43.0969097076 minutes, that is, approximately at 2:43 pm.

If the second snowplower had been dispatched at 3 pm, then in the derivation of (##), C would have been 3, hence to find the time of crash we would then need to equate

Hence, a crash would have occured at t = e² = 7.3890560989, i.e. sometime between 7 and 8 pm, and it would have been unavoidable.

Note: I attach a graph of distance vs time for the two snowplows:

We see that something peculiar happens to the distance - time profile of the second snowplow: it 'blows up' at the intersection point. This is because as the second snowplow approaches the first, the snow gets thinner and thinner, and the snowplow gets faster and faster. The portion of the two curves 'above' the profile of the first snow plow are unphysical: they can never happen, they are artifacts of the implicit functions we have used. We need to satisfy ourselves that the slopes of the curves t = (2-z)e^z at z = 1 and t = (3-z)e^z at z = 2 are non-negative; only then can we be reassured that the crash actually happened.

Indeed,

Hence we are satisfied the crash actually happens whenever the second snowplow starts.