One morning is 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, in other words its speed it inversely proportional to the depth of the snow. If the plow covered twice as much distance in the first hour as the second hour, what time did it start snowing?	Choose the correct choice from the following answers
A) 4.15 am	B) 3.20 am	C) 5.22 am	D) 6.10 am	E) 2.40 am
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 it the plow traveled twice as far in the first hour as the second: ln((x+1)/x) = ln((x+2)/(x+1))2   Exp both sides and you have (x+1)/x = ((x+2)/(x+1))2.   Solving for x you get x=(51/2-1)/2, which is the number of hours that elapsed between the time it started snowing and the snow plow left.It started snowing at (51/2-1)/2 hours before 6:00am, or 5:22:55am.
<result>C</result>
One morning is 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, in other words its speed it inversely proportional to the depth of the snow. If the plow covered twice as much distance in the first hour as the second hour, what time did it start snowing?