Question 1176132: One day snow begins to fall at a constant rate.
At noon, a snowplow starts clearing a road. The plow clears a constant volume per hour.
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At 1PM, it has cleared 2 miles.
At 2PM, it has cleared 1 more mile.
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What time did it begin to snow?
Found 3 solutions by josgarithmetic, Alan3354, ikleyn:
Answer by josgarithmetic(39620)   (Show Source):
You can put this solution on YOUR website! According to the description, the situation is a linear relationship. Noon time is x=0. The two points are (1,2) and (2,3). Already described is slope is 1 mile per hour. You finish this yourself.
When was y=0?
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! One day snow begins to fall at a constant rate.
At noon, a snowplow starts clearing a road. The plow clears a constant volume per hour.
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At 1PM, it has cleared 2 miles.
At 2PM, it has cleared 1 more mile.
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What time did it begin to snow?
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It starts snowing at 11:22:55.2
I can show how to find that if you like.
email via the TY note to see the details
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It's not linear, since the snow continues to fall.
That reduces the speed of the plow, which is a constant times the inverse of the depth of the snow.
It starts at 11:22:55.2 AM
Answer by ikleyn(52810)  (Show Source):
You can put this solution on YOUR website! .
One day snow begins to fall at a constant rate.
At noon, a snowplow starts clearing a road. The plow clears a constant volume per hour.
-------
At 1PM, it has cleared 2 miles.
At 2PM, it has cleared 1 more mile.
-------
What time did it begin to snow?
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As usual, the major point in such problem is to make a correct setup.
For it, you need a right idea and right mathematical model.
And, as usual, only a person, who is equally good in Physics and Mathematics (like me)
can do it in a right way. (You may consider it as a joke, although I am absolutely serious)
Let start making a sketch. In the Figure, I show the first 2 miles of the road (segment AB)
and the next 1 mile (segment BC)
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A (2 miles) B (1 mile) C
Let "t" denotes the time (the duration in hours) of snowing BEFORE the noon.
Let U be the rate of snowing (= the volume of the snow per mile of the road per hour).
Let V be the rate of clearing the road (= the volume of the snow taken off by the machine per hour).
Values U and V are constant in time.
During the time of t hours, the part of the road AB obtained the volume of the snow equal to
2 miles * U * t hours = 2Ut units of the snow volume.
During one hour from the noon to 1 pm, snowing added
2 miles * U * 1 hour = 2U units of the snow volume.
It was taken off by the machine in 1 hour, so we have this equation
2Ut + 2U = V (1)
It is our first equation.
During next hour from 1 pm to 2 pm, snowing added the amount of snow 2U on the part AB of the road.
On the part BC of the road, we have at 2 pm the amount of snow equal to Ut + 2U.
This amount of snow on AC, totaled 2U + (Ut + 2U), was taken off by the machine in 1 hour.
It gives us the second equation
2U + (Ut + 2U) = V (2)
Compare equations (1) and (2). Their right sides are identical --- hence, their left sides are equal
2Ut + 2U = 2U + (Ut + 2U).
Simplify
2Ut - Ut = 2U + 2U - 2U
Ut = 2U
Cancel U in both side
t = 2.
ANSWER. It started snowing 2 hours before noon, i.e. at 10 am.
Solved.
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