A cyclist leaves the town of Alphaville and heads toward Betaville. She travels at a constant speed of \(14\) km/h.
At the same time, a jogger and a walker leave Betaville and head toward Alphaville. The walker travels at a constant speed of \(6\) km/h and the jogger travels at a constant speed of \(10\) km/h.
If the cyclist passes the walker \(4\) minutes after passing the jogger, how far apart are the towns Alphaville and Betaville?
Let \(d\) be the distance, in km,
between the towns Alphaville and Betaville.
Let \(t\) be the time, in hours, until
the jogger and cyclist meet.
Using the formula distance \(=\) speed \(\times\) time, in \(t\) hours the cyclist travels \(14t\) km and the jogger travels \(10t\) km.
Between the cyclist and jogger, they travel the total distance between Alphaville and Betaville in \(t\) hours. Therefore, \(d = 14t + 10t = 24t\).
The cyclist meets the walker \(4\) minutes, or \(\frac{4}{60} = \frac{1}{15}\) hours, after meeting the jogger. Therefore, \(\left(t + \frac{1}{15}\right)\) is the time, in hours, until the cyclist meets the walker.
Again, using the formula distance \(=\) speed \(\times\) time, in \(\left(t + \frac{1}{15}\right)\) hours, the cyclist travels \(14\left(t + \frac{1}{15}\right)\) km and the walker travels \(6\left(t + \frac{1}{15}\right)\) km.
Between the cyclist and walker, they travel the total distance between Alphaville and Betaville in \(\left(t + \frac{1}{15}\right)\) hours. Therefore, \(d = 14\left(t + \frac{1}{15}\right) + 6\left(t + \frac{1}{15}\right) = 20\left(t + \frac{1}{15}\right)\).
Thus, \(d = 24t\) and \(d=20\left(t + \frac{1}{15}\right)\). Therefore, \[\begin{aligned} 24t &= 20\left(t+\frac{1}{15}\right)\\ 24t &= 20t+\frac{4}{3}\\ 4t &= \frac{4}{3}\\ t &= \frac{1}{3} \end{aligned}\] Since \(t = \frac{1}{3}\) hours, we find \(d = 24t = 24\left(\frac{1}{3}\right) = 8\) km.
Therefore, the towns of Alphaville and Betaville are \(8\) km apart.