Q. 58 - Precalculus Enhanced with Graphing Utilities

Question

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

$W=\left\{\begin{array}{ll}t& 0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}& 1.79\le v\le 20\\ 33-1.5958\left(33-t\right)& v>20\end{array}\right.$

where v represents the wind speed (in meters per second) and t represents the air temperature $\left({}^{\circ }C\right)$. Compute the wind chill for the following:

Part (a): An air temperature of $-10{}^{\circ }C$ and a wind speed of $1$ meter per second$\left(m/sec\right)$.

Part (b): An air temperature of $-10{}^{\circ }C$ and a wind speed of $5m/sec$.

Part (c): An air temperature of $-10{}^{\circ }C$ and a wind speed of $15m/sec$.

Part (d): An air temperature of $-10{}^{\circ }C$ and a wind speed of $25m/sec$.

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

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

Step-by-Step Solution

Answer

Part (a): The wind chill temperature is $-10°C$.

Part (b): The wind chill temperature is $-21.26°C$.

Part (c): The wind chill temperature is $-33.68°C$.

Part (d): The wind chill temperature is $-35.62°C$.

Part (e): The equation corresponding to the given range is $W=t$ which means the wind chill is equal to the air temperature.

Part (f): The equation corresponding to the given range is $W=33-1.5958\left(33-t\right)$  which means the wind chill factor depends only on the air temperature.

Part (a) Step 1. Given information.

Consider the given question,

$W=\left\{\begin{array}{ll}t& 0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}& 1.79\le v\le 20\\ 33-1.5958\left(33-t\right)& v>20\end{array}\right.$

Substitute $$\begin{gathered} t=-10{}^{\circ }C, \\ v=1m/sec \end{gathered}$$ in the given equation,

$$\begin{gathered} W=t \\ W=-10°C \end{gathered}$$

Part (b) Step 1. Given information.

Consider the given question,

$W=\left\{\begin{array}{ll}t& 0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}& 1.79\le v\le 20\\ 33-1.5958\left(33-t\right)& v>20\end{array}\right.$

Substitute $$\begin{gathered} t=-10{}^{\circ }C, \\ v=5m/sec \end{gathered}$$ in the given equation,

$$\begin{gathered} W=33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04} \\ W=33-\frac{\left(10.45+10\sqrt{5}-5\right)\left(33-\left(-10\right)\right)}{22.04} \\ W=33-\frac{\left(5.45+10\sqrt{5}\right)\left(33+10\right)}{22.04} \\ W=-21.26°C \end{gathered}$$

Part (c) Step 1. Given information.

Consider the given question,

$W=\left\{\begin{array}{ll}t& 0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}& 1.79\le v\le 20\\ 33-1.5958\left(33-t\right)& v>20\end{array}\right.$

Substitute $$\begin{gathered} t=-10{}^{\circ }C, \\ v=15m/sec \end{gathered}$$ in the given equation,

$$\begin{gathered} W=33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04} \\ W=33-\frac{\left(10.45+10\sqrt{15}-15\right)\left(33-\left(-10\right)\right)}{22.04} \\ W=33-\frac{\left(-4.55+10\sqrt{15}\right)\left(33+10\right)}{22.04} \\ W=-33.68°C \end{gathered}$$

Part (d) Step 1. Given information.

Consider the given question,

$W=\left\{\begin{array}{ll}t& 0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}& 1.79\le v\le 20\\ 33-1.5958\left(33-t\right)& v>20\end{array}\right.$

Substitute $$\begin{gathered} t=-10{}^{\circ }C, \\ v=25m/sec \end{gathered}$$ in the given equation,

$$\begin{gathered} W=33-1.5958\left(33-t\right) \\ W=33-1.5958\left(33-\left(-10\right)\right) \\ W=33-1.5958\left(43\right) \\ W=-35.62°C \end{gathered}$$

Part (e) Step 1. Given information.

Consider the given question,

$W=\left\{\begin{array}{ll}t& 0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}& 1.79\le v\le 20\\ 33-1.5958\left(33-t\right)& v>20\end{array}\right.$

As $0<v<1.79$, the equation for wind chill temperature W is $W=t$.

Therefore, for the given range, the wind chill is equal to the air temperature.

Part (f) Step 1. Given information.

Consider the given question,

$W=\left\{\begin{array}{ll}t& 0\le v<1.79\\ 33-\frac{\left(10.45+10\sqrt{v}-v\right)\left(33-t\right)}{22.04}& 1.79\le v\le 20\\ 33-1.5958\left(33-t\right)& v>20\end{array}\right.$

As $v>20$, the equation for wind chill temperature W is $W=33-1.5958\left(33-t\right)$.

Therefore, for the given range, the wind chill factor depends only on the air temperature.