College Algebra, Chapter 2, 2.5, Section 2.5, Problem 40

Suppose that a car is travelling on a curve that forms a circular area. Then, the force $F$ needed to keep the car from skidding is jointly proportional to the weight $w$ of the car and the square of its speed is $s$, and is inversely proportional to the radius $r$ of the curve.
a.) Write an equation that expresses the relationship.
$\displaystyle F = \frac{k w s^2}{r}$

b.) A car weighing 1600lb travels around a curve at $\displaystyle 60 \frac{\text{mi}}{\text{h}}$. The next car to round this curve weighs 2500lb and requires the same force as the first car to keep from skidding. How fast is the second car travelling?
Let $F_1, w_1$ and $s_1$, be the information about the first car. On the other hand, Let $F_2, w_2$ and $s_2$ be the infromation about the second car. Since the constant of proportionality, the radius of the curve and the force is the same on both cars.


$
\begin{equation}
\begin{aligned}
\text{Equation 1: }\\
\\
F &= \frac{k}{r} w_1 (s_1)^2 \\
\\
F &= \frac{k}{r} \left( 1600 \text{lb} \right) \left( 60 \frac{\text{mi}}{\text{h}} \right)^2\\
\\
\\
\\
\text{Equation 2: }\\
\\
F &= \frac{k}{r} w_2 (s_2)^2 \\
\\
F &= \frac{k}{r} 2500\text{lb} (s_2)^2
\end{aligned}
\end{equation}
$

Use equations 1 and 2 to solve for $s_2$
$\displaystyle \frac{Fr}{k} = \left( 1600 \text{lb} \right) \left( 60 \frac{\text{mi}}{\text{h}} \right)^2$ and $\displaystyle \frac{Fr}{k} = 2500 \text{lb}(s_2)^2$


$
\begin{equation}
\begin{aligned}
\left( 1600 \text{lb} \right) \left( 60 \frac{\text{mi}}{\text{h}} \right)^2 &= \left( 2500 \text{lb} \right)(s_2)^2 && \text{Solve for } s_2\\
\\
(s_2)^2 &= \frac{1600}{2500} \left( 60 \frac{\text{mi}}{\text{h}} \right)^2 && \text{Cancel out like terms}\\
\\
s_2 &= 48 \frac{\text{mi}}{\text{h}} && \text{The speed of the second car}
\end{aligned}
\end{equation}
$