College Algebra, Chapter 1, 1.2, Section 1.2, Problem 58

Suppose that two cyclists $A$ and $B$ are $90 mi$ apart, start riding toward each at the same time. $A$ is twice as fast as $B$. If they meet $2 h$ later, at what average speed is each cyclist traveling?

Recall that $\displaystyle \text{speed} = \frac{\text{distance}}{\text{time}}$, if we let $x$ be the speed of $B$, then $2x$ will be the speed of $A$.

Since both cyclists are traveling towards each other, the effective speed will be $2x + x = 3x$.

So,


$
\begin{equation}
\begin{aligned}

\text{time} =& \frac{\text{distance}}{\text{speed}}
\\
\\
2 =& \frac{90}{3x}
\\
\\
6x =& 90
\\
\\
x =& 15 mph

\end{aligned}
\end{equation}
$


Thus, the average speed of cyclist $A$ is $2x = 2(15) = 30 mph$ while $B$ is $15 mph$.