Two bicyclists head toward each other from the opposite ends of Main Street, which is six miles long. The first biker started at 2:05 going 12 mph. The second biker began peddling 4 minutes later at a rate of 14 mph. What time will they meet?

Two bicyclists head toward each other from the opposite ends of Main Street, which is six miles long. The first biker started at 2:05 going 12 mph. The second biker began peddling 4 minutes later at a rate of 14 mph. What time will they meet?

Detailed Explanation

The first bike got a 4/5 mile head start (12 × 4/60).

Therefore, by the time the second bike leaves, there are 51/5 miles between them (6 – 4/5).

Their combined rate of travel is 12 + 14 = 26 mph.

Let t = the number of hours the second bike travels.

26t = \(5\frac{{1} }{{5}}\)

26t = \(\frac{{26} }{{5}}\)

t = \(\frac{{26} }{{5}} ÷ \frac{{26} }{{1}}\)

t = \(\frac{{26} }{{5}} × \frac{{1} }{{26}}\)

t = \(\frac{{1} }{{5}}\)

\(\frac{{1} }{{5}}\) of an hour = 12 minutes. The second bike left at 2:09, so both bikes will meet at 2:21.

Therefore, by the time the second bike leaves, there are 51/5 miles between them (6 – 4/5).

Their combined rate of travel is 12 + 14 = 26 mph.

Let t = the number of hours the second bike travels.

26t = \(5\frac{{1} }{{5}}\)

26t = \(\frac{{26} }{{5}}\)

t = \(\frac{{26} }{{5}} ÷ \frac{{26} }{{1}}\)

t = \(\frac{{26} }{{5}} × \frac{{1} }{{26}}\)

t = \(\frac{{1} }{{5}}\)

\(\frac{{1} }{{5}}\) of an hour = 12 minutes. The second bike left at 2:09, so both bikes will meet at 2:21.