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?

Same as Charlie's, with more explanation...

Let S be the constant rate of snowfall, and say snow began falling at time t=0. The height H of the snow at any time t is

H=S*t

Let R be the constant rate of snow removal by plowing, and say the plow moves at a speed of dL/dt. Then

R = H*dL/dt
R = S*t*dL/dt
or, solving for dL,
dL = [R/(S*t)]*dt

The length L that the plow travels in a given time is simply the integral of dL over that time period. If plowing begins at time t=T, we know the plow travels twice as far during the first hour (from T to T+1 hour) as it does in the second hour (T+1 hour to T+2 hours). Therefore,

∫ [R/(S*t)]*dt from T to T+1 = 2*∫ [R/(S*t)]*dt from T+1 to T+2
R/S *ln(t) from T to T+1 = 2*R/S *ln(t) from T+1 to T+2
canceling out the R/S terms,
ln(T+1)-ln(T) = 2*[ln(T+2)-ln(T+1)]
3*ln(T+1)= 2*ln(T+2)+ln(T)
ln[(T+1)^3]=ln[T*(T+2)^2]
(T+1)^3= T*(T+2)^2
T^3+3T^2+3T+1=T^3+4T^2+4T
T^2+T-1=0
T=(-1±sqrt5)/2
T=0.6180 hours (approximately 37 minutes) or T=-1.618

Since plowing begins a positive number of minutes after snowfall begins, T>0, and therefore only the positive root makes sense. As T occurs at 6:00 a.m. , t=0 must have been 37 minutes earlier, or in other words, the snow began falling at 5:23 a.m.

Posted by Bryan on 2003-06-05 10:53:20