Question

A long distance runner starts at the beginning of a trail and runs at a rate of 4 miles per hour. Two hours later, a cyclist starts at the beginning of the trail and travels at a rate of 14 miles per hour. What is the amount of time that the cyclist travels before overtaking the runner? Do not do any rounding.

Answer 1

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.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ &------&------&------\\ Runner&d&4&t+2\\ Cyclist&d&14&t \end{array} \\\\\\ \begin{cases} d=4(t+2)\\ \boxed{d}=14t\\ -------\\ \boxed{14t}=4(t+2) \end{cases} \\\\\\ 14t=4t+8\implies 10t=8\implies t=\cfrac{8}{10} \\\\\\ t=\cfrac{4}{5}~hr\impliedby \textit{or 48 minutes}$