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anonymous
 5 years ago
One fine morning it begins to snow at a constant rate. A snow plow is a bulldozer specifically ddesigned to clear snow. At 6:00am, a snow plow starts to clear a straight road covered by snow. The plow can remove a fixed volume of snow per unit time, so that we can say that its speed is inversely proportional to the depth of the snow. If the plow covered twice as much distance in the first hour as the second hour, can you guess what time it started snowing?
i) 5:22:55 am
ii) 5:19:00 am
iii) 5:10:01 am
anonymous
 5 years ago
One fine morning it begins to snow at a constant rate. A snow plow is a bulldozer specifically ddesigned to clear snow. At 6:00am, a snow plow starts to clear a straight road covered by snow. The plow can remove a fixed volume of snow per unit time, so that we can say that its speed is inversely proportional to the depth of the snow. If the plow covered twice as much distance in the first hour as the second hour, can you guess what time it started snowing? i) 5:22:55 am ii) 5:19:00 am iii) 5:10:01 am

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0answer is (i) 5:22:55 am Let x be time it snows before 6am snow is falling at constant rate R depth of snow is proportional to amount of time its been snowing depth = Rt speed is inversely proportional to depth, s = k/Rt distance = speed*time since speed is constantly changing use integration to find distance\[\frac{k}{R}\int\limits_{x}^{x+1}\frac{1}{t} dt = 2\frac{k}{R}\int\limits_{x+1}^{x+2}\frac{1}{t} dt\] 1st integral finds distance for first hour, 2nd integral find distance for 2nd hour the k/R will cancel leaving \[\ln \frac{x+1}{x} = 2\ln \frac{x+2}{x+1}\] make both sides power of e\[\frac{x+1}{x} = \frac{(x+2)^{2}}{(x+1)^{2}}\] cross multiply and simplify\[x^{2} +x 1 =0\] quadratic formula results in\[x = \frac{1\pm \sqrt{5}}{2}\] x >0 because it represents positive time x = .618 hours = 37.08 min
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