Chapter 3.6, Problem 61PS

Solution

i.

To state: The wind speed that corresponds to the Beaufort number $B=0$ .

The heights that the basketball player and the kangaroo jumped are $2.6244\text{m}\text{and}5.0176\text{m}$ respectively.

Given information:

The given function is $B=1.69\sqrt{s+4.25}-3.55$ .

The Beaufort wind scale was devised to measure wind speed. The Beaufort numbers B, which range from 0 to 12, can be modeled by $B=1.69\sqrt{s+4.25}-3.55$ where s is the speed (in miles per hour) of the wind.

  

Calculation:

First, solve for s .

Write the given equation:

  $B=1.69\sqrt{s+4.25}-3.55$ .

Add 3.55 on both sides of the equation,

  $\begin{array}{l}B+3.55=1.69\sqrt{s+4.25}-3.55+3.55\\ B+3.55=1.69\sqrt{s+4.25}\end{array}$

Divide both sides of the equation by $1.69$ ,

  $\begin{array}{l}\frac{B+3.55}{1.69}=\frac{1.69\sqrt{s+4.25}}{1.69}\\ \frac{B+3.55}{1.69}=\sqrt{s+4.25}\end{array}$

Square both sides of the equation,

  $\begin{array}{c}{\left(\frac{B+3.55}{1.69}\right)}^2={\left(\sqrt{s+4.25}\right)}^2\\ {\left(\frac{B+3.55}{1.69}\right)}^2=s+4.25\end{array}$

Now subtract $4.25$ from both sides of the equation,

  $\begin{array}{l}{\left(\frac{B+3.55}{1.69}\right)}^2-4.25=s+4.25-4.25\\ {\left(\frac{B+3.55}{1.69}\right)}^2-4.25=s\end{array}$

Now substitute 0 for B to find the height of the jump by basketball player,

  $\begin{array}{c}{\left(\frac{0+3.55}{1.69}\right)}^2-4.25=s\\ {\left(2.101\right)}^2-4.25\approx s\\ 4.414-4.25\approx s\\ 0.164\approx s\end{array}$

Therefore, wind speed that corresponds to the Beaufort number $B=0$ is 0.164 miles per hour.

b.

To state: The wind speed that corresponds to the Beaufort number $B=12$ .

The new heights that the basketball player and the kangaroo jumped are $\text{10}\text{.4976}\text{m}\text{and}20.0704\text{m}$ respectively.

Given information:

The equation from part (a) is ${\left(\frac{0+3.55}{1.69}\right)}^2-4.25=s$ .

Explanation:

Now substitute 12 for B to find the height of the jump by basketball player,

  $\begin{array}{c}{\left(\frac{12+3.55}{1.69}\right)}^2-4.25=s\\ {\left(\frac{15.55}{1.69}\right)}^2-4.25\approx s\\ 84.66-4.25\approx s\\ 80.4\approx s\end{array}$

Therefore, wind speed that corresponds to the Beaufort number $B=12$ is 80.4 miles per hour.

c.

To state: An inequality that describes the range of wind speeds represented by the Beaufort model.

The resultant inequality is $0.162\le x\le 80.4$ .

Given information:

The statement says, “Write an inequality that describes the range of wind speeds represented by the Beaufort model?”

Explanation:

The wind speed is 0.162 miles per hour for the lowest value of B and 80.4 miles per hour for the highest value of B as a result.

Specifically, the lowest wind speed is 0.162 miles/h and the highest wind speed is 80.4 miles/h.

As a result, the inequality representing the wind speed range is $0.162\le x\le 80.4$ .