Chapter 3.5, Problem 39PS

Solution

$\left(a\right)$

To write an equation that gives $v_t$ as function of $A$ for a skydiver who weighs $165$ pounds.

  $v_t=\frac{433}{\sqrt{A}}$

Given:

Under certain condition, a skydiver’s terminal velocity $v_t$ (in feet per second) is given by following equation:

  $v_t=33.7\sqrt{\frac{W}{A}}$

where $W$ is the weight of the skydiver (in pounds) and $A$ is the skydiver’s cross-sectional area (in square feet) and the skydiver can vary their cross-sectional surface area by changing position as they fall.

Calculation:

Since the skydiver’s terminal velocity $v_t$ (in feet per second) is given by following equation:

  $v_t=33.7\sqrt{\frac{W}{A}}$

To write an equation for the skydiver who weighs $165$ pounds, substitute $165$ for $W$ in the above equation:

  $\begin{array}{l}{v}_t=33.7\sqrt{\frac{165}{A}}\\ \Rightarrow =33.7\times \frac{12.84}{\sqrt{A}}\\ \Rightarrow =\frac{433}{\sqrt{A}}\end{array}$

$\left(b\right)$

To make table of values for the equation from part $\left(a\right)$ .

  

Given:

Under certain condition, a skydiver’s terminal velocity $v_t$ (in feet per second) is given by following equation:

  $v_t=33.7\sqrt{\frac{W}{A}}$

where $W$ is the weight of the skydiver (in pounds) and $A$ is the skydiver’s cross-sectional area (in square feet) and the skydiver can vary their cross-sectional surface area by changing position as they fall.

Calculation:

Use the following equation from $\left(a\right)$ part to make the table:

  $v_t=\frac{433}{\sqrt{A}}$

  $\begin{array}{l}\text{When}A=1\\ v_t=\frac{433}{\sqrt{1}}=433\end{array}$

  $\begin{array}{l}\text{When}A=4:\\ v_t=\frac{433}{\sqrt{4}}\\ \Rightarrow =\frac{433}{2}\\ \Rightarrow =216.5\end{array}$

  $\begin{array}{l}\text{When}A=9:\\ v_t=\frac{433}{\sqrt{9}}\\ \Rightarrow =\frac{433}{3}\\ \Rightarrow =144.3\end{array}$

  $\begin{array}{l}\text{When}A=16:\\ v_t=\frac{433}{\sqrt{16}}\\ \Rightarrow =\frac{433}{4}\\ \Rightarrow =108.2\end{array}$

$\left(c\right)$

Use the table of values in part $\left(b\right)$ to graph the equation.

Given:

Under certain condition, a skydiver’s terminal velocity $v_t$ (in feet per second) is given by following equation:

  $v_t=33.7\sqrt{\frac{W}{A}}$

where $W$ is the weight of the skydiver (in pounds) and $A$ is the skydiver’s cross-sectional area (in square feet) and the skydiver can vary their cross-sectional surface area by changing position as they fall.

Graph: