Gabi and Silvio are training for a cycling race. They live on the same street, but Silvio’s house is \(2\) km east of Gabi’s. On Sunday morning at \(7\) a.m. they each start biking east from their house. If Gabi bikes at a constant speed of \(24\) km/h and Silvio bikes at a constant speed of \(18\) km/h, at what time will Gabi catch up to Silvio?
For the first two solutions we will use the formula: \(\text{time}=\frac{\text{distance}}{\text{speed}}\).
For the third solution we will use the formula: \(\text{distance}=\text{speed} \times \text{time}\).
Solution 1
Since Gabi bikes at \(24\) km/h and Silvio bikes at \(18\) km/h, then Gabi gains \(6\) km/h on Silvio.
Since Silvio starts \(2\) km east of of Gabi, then it takes Gabi \(\frac{2}{6}=\frac{1}{3}\) of an hour or \(\frac{1}{3} \times 60 = 20\) minutes to catch up to Silvio. Since they started biking at \(7\) a.m., Gabi will catch up to Silvio at \(7\):\(20\) a.m.
Solution 2
Silvio bikes at \(18\) km/h or \(\frac{18}{60} = \frac{3}{10}\) km/min. Gabi bikes at \(24\) km/h or \(\frac{24}{60} = \frac{2}{5}\) km/min. Therefore Gabi gains \(\frac{2}{5} - \frac{3}{10} = \frac{1}{10}\) km/min on Silvio.
Since Silvio started \(2\) km east of Gabi, then it takes Gabi \(2 \div \frac{1}{10} = 20\) minutes to catch Silvio. Since they started biking at \(7\) a.m., Gabi will catch up to Silvio at \(7\):\(20\) a.m.
Solution 3
Suppose it takes \(t\) hours for Gabi to catch up to Silvio. Then Silvio has biked \(18 \text{ km/h} \times t\text{ h} = 18t \text{ km}\), and Gabi has biked \(24 \text{ km/h} \times t\text{ h} = 24t \text{ km}\).
Since Silvio starts \(2\) km east of Gabi, then when they meet, Gabi will have travelled \(2\) km further than Silvio. That is, \[\begin{aligned} 24t& = 18t + 2\\ 6t &= 2\\ t &= \tfrac{1}{3} \end{aligned}\] Therefore, it takes Gabi \(\frac{1}{3}\) of an hour, or \(20\) minutes to catch up to Silvio. Since they started biking at \(7\) a.m., Gabi will catch up to Silvio at \(7\):\(20\) a.m.