Q. 57 - Precalculus Enhanced with Graphing Utilities

Question

Wind Chill The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is

$$\begin{gathered} W=\{t 0\le v<1.79 \\ 33-\frac{(10.45+10\sqrt{v}-v)(33-t)}{22.04},1.79\le v\le 20 \\ 33-1.5958(33-t),v>20 \end{gathered}$$

where $v$ represents the wind speed (in meters per second) and $t$ represents the air temperature . Compute the wind chill for the following:

(a) An air temperature of $10°\mathrm{C}$ and a wind speed of 1 meter per second $(\mathrm{m}/\sec )$

(b) An air temperature of $10°\mathrm{C}$ and a wind speed of $5 \mathrm{m}/\sec$

(c) An air temperature of $10°\mathrm{C}$ and a wind speed of $15 \mathrm{m}/\sec$

(d) An air temperature of $10°\mathrm{C}$ and a wind speed of $25 \mathrm{m}/\sec$

(e) Explain the physical meaning of the equation corresponding to $0\le v<1.79$

(f) Explain the physical meaning of the equation corresponding to $v>20$.

Step-by-Step Solution

Answer

The values are $10,3.98,-2.66,-3.7$ respectively and wind speed is equal to chill temperature

Part (a) Step 1: Given information

Given the air temperature of ${10}^{\circ }C$ and wind speed of $1m/s$

Part (a) Step 2: Calculating the values

As $v=1$ is between 0 and $1.79$ we have

$W=t$

Substitute $10$ for $t$

$\Rightarrow W=10$

Part (b) Step 1: Given information

Given the air temperature of ${10}^{\circ }C$ and wind speed of $15m/s$

Part (b) Step 2: Substitute t = 10 , v = 5 and calculate the value

Here $v=5$ is between $1.79$ and $20$ , so we have

$$\begin{gathered} W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04} \\ W=33-\frac{(10.45+10\sqrt{5}-5)·(33-10)}{22.04} \\ W=33-\frac{(5.45+22.36)·23}{22.04} \\ W=33-\frac{27.81·23}{22.04} \\ W=33-29.02 \\ W=3.98 \end{gathered}$$Substituting, we get

Here $v=5$ is between $1.79$ and $20$, so we have

$W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04}$

Part (c) Step 1: Given information

Given the air temperature of ${10}^{\circ }C$ and wind speed of $25m/sec$

Part (c) Step 2: Substitute t = 10 , v = 15 and calculate the value

Substituting, we get

$v=15$ is between $1.79$ and $20$ , so we have

$W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04}$

$$\begin{gathered} W=33-\frac{(10.45+10\sqrt{v}-v)·(33-t)}{22.04} \\ W=33-\frac{(10.45+10\sqrt{15}-15)·(33-10)}{22.04} \\ W=33-\frac{(-4.55+38.72)·23}{22.04} \\ W=33-\frac{34.17·23}{22.04} \\ W=33-35.66 \\ W=-2.66 \end{gathered}$$

Part (d) Step 1: Given information

Given the air temperature of ${10}^{\circ }C$ and wind speed of $25m/sec$

Part (d) Step 2: Substitute t = 10 and solve

Calculating, we get

As $25>20$

$W=33-1.5958(33-t)$

$$\begin{gathered} W=33-1.5958(33-10) \\ W=33-36.7 \\ W=-3.7 \end{gathered}$$

Part (e) Step 1: Given information

Given the equation corresponding to $0\le v<1.79$

Part (e) Step 2: Checking the required equation

Here for $0\le v<1.79$ we have

$W=t$

Therefore, the wind chill is equal to the air temperature for $0\le v<1.79$

Part (f) Step 1: Given information

Given the equation corresponding to $v>20$

Part (f) Step 2: Checking the required equation

Here for 

From the function for $v>20$ we have

$W=33-1.5958(33-t)$

Therefore, for $v>20$ the wind chill depends only on the air temperature.