One morning it starts to snow at a constant rate. Later, at 6:00am, a snow plow sets out to clear a straight street. The plow can remove a fixed volume of snow per unit time.
If the plow covered twice as much distance in the first hour as the second hour, what time did it start snowing?
Since the plow can remove a fixed volume of snow per unit time, its speed is inversely proportional to the depth of the snow.
Let the depth of snow at time t to be t units. The speed of the plow at time t will be 1/t. Define t=0 as the time it started snowing and t=x the time the plow started.
The distance covered in the first hour is the integral from x to x+1 of 1/t dt. The antiderivative of 1/t is ln(t) so the total distance covered in the first hour is
ln((x+1)/x).
By the same reasoning, the distance covered in the second hour in ln((x+2)/(x+1)).
Using the fact that the plow traveled twice as far in the first hour as the second:
ln((x+1)/x) = 2 ln((x+2)/(x+1))
ln((x+1)/x) = ln((x+2)/(x+1))²
Expand both sides (using ln a=ln b → a=b) and you have:
(x+1)/x = ((x+2)/(x+1))².
Solving for x you get:
x=(√5-1)/2
This is the number of hours that elapsed between the time it started snowing and when the snow plow began.
(√5-1)/2 is approximately 0.61803398874989 hours.
Multiply by 60 to get a figure of about 37.08203932499369 minutes.
Multiply the decimal part of this by 60 again to get 4.92235949962, or about five seconds.
So, it started snowing at (√5-1)/2 hours, or 37 minutes 5 seconds, before six o'clock, which is 5:22:55 am.
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