One morning it began to snow very hard in the mountains and continued snowing st

Question

One morning it began to snow very hard in the mountains and continued snowing steadily throughout the day. A snowplow set out at 8:00am to clear the highway, clearing 5 km by 11:00am, and an additional 2 km by 1:00pm. At what time did it start snowing? To solve this problem, you must make two physical assumptions concerning the rate at which it is snowing at the rate at which the snowplow can clear the road. Since it is snowing steadily, it is reasonable to assume that it is snowing at a constant rate. From the data given (and from our experience), the deeper the snow, the slower the snowplow moves. With this in mind, assume that the rate (in km/h) at which a snowplow can clear a road is inversely proportional to the depth of the snow.

Explanation / Answer

some very good links for your solution http://www.slideshare.net/mcirish3/snow-plow-know-how sample question One morning it began to snow very hard and continued to snow steadily through the day. A snowplow set out at 8:00 A.M. to clear a road, clearing 2 miles by 11:00 A.M. and an additional mile by 1:00 P.M. At what time did it start snowing. (You may assume that it was snowing at a constant rate and that the rate at which the snowplow could clear the road was inversely proportional to the depth of the snow. Say t = 0 at 8:00 a.m. ; Define function snow as ds/dt = C ==> ds = C dt ==> s = Ct + D ; C ? 0 The rate the snow plough clears the road dy/dt = k/s = k/(Ct + D) dy = k/(Ct + D) dt Integrate y(t) = k/C ln(Ct + D) + E We have y(0) = 0 ; at 8:00 , there is nothing cleared ; Also, we conveniently choose E = 0 , since we assume the initial point is the point the plough starts clearing the snow. Then, since k ? 0 , 0 = k/C ln ( D ) + 0 ==> ln D = 0 ==> D = 1 The initial conditions 11:00 a.m. is t = 3 y(3) = k/C ln (3C + 1) (i) y(3) = 2 mi 1:00 p.m. is t = 5 y(5) = k/C ln (5C + 1) (i) y(5) = 3 mi Dividing (i) by (ii) 2/3 = ln (3C + 1) / ln (5C + 1) 2 ln (5C + 1) = 3 ln (3C + 1) (5C + 1)² = (3C + 1)³ For this equation use some software (eg here http://www.wolframalpha.com/input/?i=%28… ) Choosing the right value ( C ? 0 and C > 0) C ˜ 0.159 Then s(t) = 0.159t + 1 ; Finding the moment when s(t) = 0 0 = 0.159t + 1 t = -1/0.159 ˜ - 6.28931 hours So the snow started at 8 - 6.28931 ˜ 1.7107 ˜ 1:42:38 a.m

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