(b) The first snow of the season begins to fall during the night. The depth of the 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, $v \, \text{km/h}$ , of the snow plough is inversely proportional to the depth of snow. (This means $v=\frac{A}{h}$ where $A$ is a constant.) Let $x \, \text{km}$ 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 $6$ am. (i) Explain carefully why, for $t \geq T$ , $6$ $0$ , where $6$ $1$ is a constant. (ii) In the period from $6$ am to $6$ $3$ am the snow plough clears $6$ $4$ of road, but it takes a further $6$ $5$ hours to clear the next kilometre. At what time did it begin snowing?

View step-by-step help

Home

Math Problems

Algebra 2

Transformations of absolute value functions

Full solution

Q. (b) The first snow of the season begins to fall during the night. The depth of the 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,

v \, \text{km/h}

, of the snow plough is inversely proportional to the depth of snow. (This means

v=\frac{A}{h}

where

A

is a constant.) Let

x \, \text{km}

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

6

am. (i) Explain carefully why, for

t \geq T

6

0

, where

6

1

is a constant. (ii) In the period from

6

am to

6

3

am the snow plough clears

6

4

of road, but it takes a further

6

5

hours to clear the next kilometre. At what time did it begin snowing?

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 $h = kt$ $0$ and $h = kt$ $1$ are constants, we can write $h = kt$ $2$ , where $h = kt$ $3$ is a new constant that combines $h = kt$ $0$ and $h = kt$ $1$ .
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 $T$ : $\newline$ $1$ . $k' [\ln(T + 2) - \ln(T)] = 1$ $\newline$ $2$ . $k' [\ln(T + 5.5) - \ln(T + 2)] = 1$ $\newline$ 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 $k'$ : $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $\newline$ This simplifies to: $\newline$ $\ln(T + 2) - \ln(T) = \ln(T + 5.5) - \ln(T + 2)$ $\newline$ Using the properties of logarithms, we can further simplify: $\newline$ $\ln\left(\frac{T + 2}{T}\right) = \ln\left(\frac{T + 5.5}{T + 2}\right)$
Final Time Calculation: Divide the first equation by the second to eliminate $k'$ : $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $\newline$ This simplifies to: $\newline$ $\ln(T + 2) - \ln(T) = \ln(T + 5.5) - \ln(T + 2)$ $\newline$ Using the properties of logarithms, we can further simplify: $\newline$ $\ln\left(\frac{T + 2}{T}\right) = \ln\left(\frac{T + 5.5}{T + 2}\right)$ Since the natural logarithm is a one-to-one function, we can equate the arguments of the logarithms: $\newline$ $\frac{T + 2}{T} = \frac{T + 5.5}{T + 2}$ $\newline$ Cross-multiply to solve for $T$ : $\newline$ $T(T + 5.5) = (T + 2)^2$ $\newline$ Expand and simplify: $\newline$ $T^2 + 5.5T = T^2 + 4T + 4$ $\newline$ Subtract $T^2$ from both sides and combine like terms: $\newline$ $5.5T - 4T = 4$ $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $0$ $\newline$ Divide by $1$ . $5$ to solve for $T$ : $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $2$ $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $3$ or approximately $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $4$ hours.
Final Time Calculation: Divide the first equation by the second to eliminate $k'$ : $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $\newline$ This simplifies to: $\newline$ $\ln(T + 2) - \ln(T) = \ln(T + 5.5) - \ln(T + 2)$ $\newline$ Using the properties of logarithms, we can further simplify: $\newline$ $\ln\left(\frac{T + 2}{T}\right) = \ln\left(\frac{T + 5.5}{T + 2}\right)$ Since the natural logarithm is a one-to-one function, we can equate the arguments of the logarithms: $\newline$ $\frac{T + 2}{T} = \frac{T + 5.5}{T + 2}$ $\newline$ Cross-multiply to solve for $T$ : $\newline$ $T(T + 5.5) = (T + 2)^2$ $\newline$ Expand and simplify: $\newline$ $T^2 + 5.5T = T^2 + 4T + 4$ $\newline$ Subtract $T^2$ from both sides and combine like terms: $\newline$ $5.5T - 4T = 4$ $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $0$ $\newline$ Divide by $1$ . $5$ to solve for $T$ : $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $2$ $\newline$ $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $3$ or approximately $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $4$ 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 $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $8$ hours, then the snow began falling around $3$ : $20$ am (since $\frac{\ln(T + 2) - \ln(T)}{\ln(T + 5.5) - \ln(T + 2)} = 1$ $9$ hours is approximately $40$ minutes).