(b) The first snow of the season begins to fall during the night. The depth of the 6 snow, h, increases at a constant rate through the night and the following day. At 6 am a snow plough begins to clear the road of snow. The speed, vkm//h, of the snow plough is inversely proportional to the depth of snow. (This means v=(A)/(h) where A is a constant.) Let xkm be the distance the snow plough has cleared and let t be the time in hours from the beginning of the snowfall. Let t=T correspond to 6am. (i) Explain carefully why, for t = T, (dx)/(dt)=(k)/(t), where k is a constant. (ii) In the period from 6 am to 8 am the snow plough clears 1km of road, but it takes a further 3.5 hours to clear the next kilometre. At what time did it begin snowing?

Question

(b) The first snow of the season begins to fall during the night. The depth of the 6 snow,  h, increases at a constant rate through the night and the following day. At 6 am a snow plough begins to clear the road of snow. The speed,  vkm//h, of the snow plough is inversely proportional to the depth of snow. (This means  v=(A)/(h) where  A is a constant.) Let  xkm be the distance the snow plough has cleared and let  t be the time in hours from the beginning of the snowfall. Let  t=T correspond to  6am. (i) Explain carefully why, for  t >= T,  (dx)/(dt)=(k)/(t), where  k is a constant. (ii) In the period from 6 am to 8 am the snow plough clears  1km of road, but it takes a further 3.5 hours to clear the next kilometre. At what time did it begin snowing?

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Given Information: We are given that the speed of the snow plough, $ v $, is inversely proportional to the depth of the snow, $ h $, which means $ v = \frac{A}{h} $, where $ A $ is a constant. We also know that the depth of the snow increases at a constant rate, so $ h $ is directly proportional to $ t $, the time since the beginning of the snowfall. Therefore, we can write $ h = kt $, where $ k $ is a constant of proportionality.Speed and Depth Relation: Since $ v = \frac{A}{h} $ and $ h = kt $, we can substitute $ h $ in the expression for $ v $ to get $ v = \frac{A}{kt} $. The distance $ x $ that the snow plough clears is the integral of its speed over time, so $ \frac{dx}{dt} = v $. Therefore, $ \frac{dx}{dt} = \frac{A}{kt} $. We can see that $ \frac{dx}{dt} $ is inversely proportional to $ t $, and since $ A $ and $ k $ are constants, we can write $ \frac{dx}{dt} = \frac{k'}{t} $, where $ k' $ is a new constant that combines $ A $ and $ k $.Distance Cleared Calculation: We are told that the snow plough clears 1 km of road from 6 am to 8 am. This is a 2-hour period, so we can set up an integral from $ T $ to $ T + 2 $ (where $ T $ is the time corresponding to 6 am in hours since the beginning of the snowfall) to find the distance cleared: $ \int_{T}^{T+2} \frac{k'}{t} dt = 1 $.First Integral Calculation: Solving the integral, we get $ k' \ln(t) $ evaluated from $ T $ to $ T + 2 $. This gives us $ k' [\ln(T + 2) - \ln(T)] = 1 $.Second Integral Calculation: We also know that it takes a further 3.5 hours to clear the next kilometre. So, we set up another integral from $ T + 2 $ to $ T + 2 + 3.5 $ to find the distance cleared: $ \int_{T+2}^{T+5.5} \frac{k'}{t} dt = 1 $.Equations Setup: Solving this integral, we get $ k' \ln(t) $ evaluated from $ T + 2 $ to $ T + 5.5 $. This gives us $ k' [\ln(T + 5.5) - \ln(T + 2)] = 1 $.Equations Solution: We now have two equations with two unknowns, $ k' $ and $ T $: 1. $ k' [\ln(T + 2) - \ln(T)] = 1 $ 2. $ k' [\ln(T + 5.5) - \ln(T + 2)] = 1 $  We can solve these equations simultaneously to find the values of $ k' $ and $ T $.Final Time Calculation: Divide the first equation by the second to eliminate $ k' $: $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$  This simplifies to: $\ln(T + 2) - \ln(T) = \ln(T + 5.5) - \ln(T + 2)$  Using the properties of logarithms, we can further simplify: $\ln\left(\frac{T + 2}{T}\right) = \ln\left(\frac{T + 5.5}{T + 2}\right)$Since the natural logarithm is a one-to-one function, we can equate the arguments of the logarithms: $\frac{T + 2}{T} = \frac{T + 5.5}{T + 2}$  Cross-multiply to solve for $ T $: $T(T + 5.5) = (T + 2)^2$  Expand and simplify: $T^2 + 5.5T = T^2 + 4T + 4$  Subtract $ T^2 $ from both sides and combine like terms: $5.5T - 4T = 4$  $1.5T = 4$  Divide by 1.5 to solve for $ T $: $T = \frac{4}{1.5}$  $T = \frac{8}{3}$ or approximately $2.67$ hours.Since $ T $ corresponds to 6 am, we need to subtract $ T $ from 6 am to find the time it began snowing. If $ T $ is approximately $ 2.67 $ hours, then the snow began falling around 3:20 am (since $ 0.67 $ hours is approximately 40 minutes).

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