recall your d = rt, distance = rate * time.
the runner takes off and goes at 4mph.
the cyclist takes off 2 hours later, and goes 14mph.
now, when they both meet, namely the cyclist comes from behind and meets ahead the runner, the distances both of them travelled, is say "d" miles for both, since both of them are "d" miles from the starting point.
if by them the cyclist has been going for say "t" hours, we know she took 2 hours later, so by the the runner has been running for "t + 2" hours then.
[tex]\bf \begin{array}{lcccl}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Runner&d&4&t+2\\
Cyclist&d&14&t
\end{array}
\\\\\\
\begin{cases}
d=4(t+2)\\
\boxed{d}=14t\\
-------\\
\boxed{14t}=4(t+2)
\end{cases}
\\\\\\
14t=4t+8\implies 10t=8\implies t=\cfrac{8}{10}
\\\\\\
t=\cfrac{4}{5}~hr\impliedby \textit{or 48 minutes}[/tex]
