Logic Puzzles

Consider the snowplow's speed and timing.

Let’s set up some notation for the problem:

t = time measured from noon (hours)
x(t) = distance the snowplow has traveled (miles)
h(t) = height of the snow (in inches)
b = hours before noon when it started snowing
k1 = rate of snowfall (inches/hour)

First, we know the snow is falling at a constant rate, so the height of snow is a linear function. At b hours before noon there is no snow, so by noon k1b inches have fallen. Thereafter, we get k1t inches per hour. So we have the equation:

h(t) = k1b + k1t = k1(t + b)

The rate the snowplow moves, its speed x‘(t), is inversely proportional to the height of snow. So for some constant k2, we have:

h(t) x’(t) = k2

x’(t) = k2/h(t)

x’(t) = k2/(k1(t + b))

x’(t) = k3/(t + b)

where k3 = k1/k2

Now we have the speed of the snowplow expressed as a separable differential equation, so we can solve it by integrating both sides.

x(t) = k3 ln(t + b) + C1

The key is solving just for the distances traveled by evaluating the anti-derivative at those points.

We use the given information. After 1 hour the snowplow has traveled 2 miles, so x(1) = 2.

x(1) = 2 = k3 ln(t + b) + C1 evaluated from t = 0 to t = 1

x(1) = 2 = k3 ln(1 + b) – k3 ln(0 + b)

x(1) = 2 = k3 ln((b + 1)/b)

And then in the next hour, the snowplow travels 1 more mile, so x(2) – x(1) = 1.

x(2) – x(1) = 1 = k3 ln(t + b) + C1 evaluated from t = 1 to t = 2

1 = k3 ln(2 + b) – k3 ln(1 + b)

1 = k3 ln[(2 + b)/(1 + b)]

We can multiply this equation above by 2, and then set it equal to the equation for x(1) = 2.

k3 ln((b + 1)/b) = 2k3 ln[(2 + b)/(1 + b)]

ln((b + 1)/b) = ln[(2 + b)/(1 + b)]2

In order for these to be equal, we set their arguments equal to each other.

(b + 1)/b = (2 + b)2/(1 + b)2

(b + 1)3 = b(2 + b)2

b3 + 3b2 + 3b + 1 = 4b + 4b2 + b3

b2 + b – 1 = 0

We can then solve this using the quadratic formula:

b = (-1 + √5)/2 ≈ 0.618, or

b = (-1 – √5)/2 ≈ -1.618

Recall that b is the time in hours before noon, so we want a positive value for b. This is the first solution, which is approximately 0.618 hours, so b is about (0.618)(60) = 37 minutes.

So the snow started about 37 minutes before noon at 11:23 a.m.