Answer:
2 hours
Explanation:
[tex]\text{Speed}=\frac{Dis\tan ce}{Time}[/tex]The first bicyclist rides at a speed of 8 mph. Therefore:
[tex]\begin{gathered} 8=\frac{d}{t} \\ \implies d=8t \end{gathered}[/tex]One hour later, the second bicyclist leaves and rides at a speed of 12 mph.
Therefore, the time of the second bicyclist = (t-1) hours.
Therefore:
[tex]\begin{gathered} 12=\frac{d}{t-1} \\ \implies d=12(t-1) \end{gathered}[/tex]Since the second bicyclist will catch up to the first bicyclist, the distance traveled will be the same.
So:
[tex]\begin{gathered} 8t=12(t-1) \\ 8t=12t-12 \\ 8t-12t=-12 \\ -4t=-12 \\ \frac{-4t}{-4}=\frac{-12}{-4} \\ t=3\text{ hours} \end{gathered}[/tex]Therefore, the second bicyclist will have traveled for:
(t-1) = (3-1) =2 hours.
