Question

The Snowplow Problem

To apply the techniques discussed in this chapter to real-world problems, it is neces-

sary to translate these problems into questions that can be answered mathematically.

The process of reformulating a real-world problem as a mathematical one often requires

making certain simplifying assumptions. To illustrate this, consider the following snow-

plow problem:

One morning it began to snow very hard and continued snowing steadily

throughout the day. A snowplow set out at 9:00 am to clear a road,

clearing 2 mi by 11:00 am and an additional mile by 1:00 pm. At what

time did it start snowing?

To solve this problem, you can make two physical assumptions concerning the rate

at which it is snowing and the rate at which the snowplow can clear the road. Because

it is snowing steadily, it is reasonable to assume it is snowing at a constant rate. From

the data given (and from our experience), the deeper the snow, the slower the snowplow

moves. With this in mind, assume that the rate (in mph) at which a snowplow can clear

a road is inversely proportional to the depth of the snow.

please show all work

Answer 1

Let us assume the following things:

1. At time t = 0 (in hours), the snowplow started operation. Hence it will have cleared 2 miles by t = 2, and 3 miles by t = 4.
2. At some t = t₀ < 0, the snow started falling. t₀ is a negative number as the snow must have started falling before the plower started (as otherwise the inverse dependency of the velocity of the plower on height of fallen snow is nonsensical).

Using these assumptions, the height h (in appropriate units e.g. metres, it does not matter which unit) of snow as a function of t is given by

here r is the rate of falling of snow, assumed constant (in units of length per hour).

The distance covered by the plow, x (in miles) is inversely proportional to the depth of the snow, and hence satisfies the equation

where k is another appropriate constant with appropriate physical units of mi² / hour. Using the form for h(t) we get

where γ = k/r is constant. Integrating ($), we get

where C is a constant of integration. (#) hence has three unknowns: γ, t₀ and C. And we have three 'boundary conditions' given:

Plugging in the first condition, we get

Hence (#) modifies to

Plugging in the second third conditions into (##) we get

Eliminating γ between these two equations we get

Simplifying the cubic, we get

We can only take the negative value of x as the true solution (as t₀ must be negative, hence we get)

Hence it began to snow approximately 1 hour and (0.236067977)*60, i.e. 14.16407862 (approximately 14) minutes before 9 am, that is at approximately 7:46 am (to be more precise, you can work out in seconds too! )