Chapter 1.5, Problem 42PS

Solution

a.

To find: The speed of a racing cyclist who experiences $5$ pounds of sir resistance

The speed of a racing cyclist who experiences $5$ pounds of sir resistance is equal to $s\approx 24.56\text{mph}$ .

Given:

The air resistance $R$ (in pounds) on a racing cyclist is given by the equation $R=0.00829s^2$ where $s$ is the bicycle's speed (in miles per hour).

Calculation:

Put $R=5$ in $R=0.00829s^2$ ,

  $\begin{array}{c}5=0.00829s^2\\ s^2=\frac{5}{0.00829}\\ s=\pm \sqrt{\frac{5}{0.00829}}\end{array}$

As $s<0$ ,

  $\begin{array}{c}s=\sqrt{603.136}\\ s\approx 24.56\end{array}$

Therefore,

The speed of racing cyclist is $s\approx 24.56\text{mph}$

b.

To explain: What happens to the air resistance if the cyclist's speed doubles?

The air resistance will be four times as that of initial speed.

Given:

The air resistance $R$ (in pounds) on a racing cyclist is given by the equation $R=0.00829s^2$ where $s$ is the bicycle's speed (in miles per hour).

Explanation:

If the speed of the cyclist doubles, then

  $\begin{array}{c}R=0.00829{\left(2s\right)}^2\\ =0.00829\left(4s^2\right)\\ =4\left[0.00829s^2\right]\end{array}$

The air resistance will be four times as that of initial speed.