Chapter 3.6, Problem 4GP

Solution

To calculate: The air pressure at the center of a hurricane when the mean sustained wind velocity is 48.3 meters per second. The function used in example 2 is $v\left(p\right)=6.3\sqrt{1013-p}$ .

The air pressure at the center of a hurricane is 954.17 millibars.

Given information:

The given function is $v\left(p\right)=6.3\sqrt{1013-p}$ .

Mean sustained wind velocity is $v\left(p\right)=48.3m/s$

Formula used: Raise each side of the equation to the same power to eliminate the radical.

Calculation:

Write the original function:

  $v\left(p\right)=6.3\sqrt{1013-p}$

Substitute the value of $v\left(p\right)$ as 48.3,

  $48.3m/s=6.3\sqrt{1013-p}$ .

Divide both sides by 6.3 to isolate the radical on one side of the equation,

  $\begin{array}{c}\frac{48.3m/s}{6.3}=\frac{6.3\sqrt{1013-p}}{6.3}\\ 7.67m/s=\sqrt{1013-p}\end{array}$

Square each side to eliminate the radical and simplify,

  $\begin{array}{c}{\left(\sqrt{1013-p}\right)}^2={\left(7.67m/s\right)}^2\\ 1013-p=58.83m^2/s^2\end{array}$

Subtract 1013 from both sides,

  $\begin{array}{c}1013-p-1013=58.83-1013\\ -p=-954.17\\ p=954.17\end{array}$

Therefore, the pressure at the center of the hurricane is $p=954.17$ millibars.