<PRE><B>Question 252492</B>
Two cyclist start biking from a trail&#39;s start 3 hours apart,
 The second cyclist travels @ 10mph and starts 3 hours after the first cyclist who is traveling @ 6mph.
 How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
:
Let t = travel time of the 2nd cyclist
then
(t+3) = travel time of the 1st cyclist
:
When the 2nd catches the 1st, they will have traveled the same distance
:
Write a distance equation: dist = speed * time
:
2nd cyclist dist = 1st cyclist dist
10t = 6(t+3)
10t = 6t + 18
10t - 6t = 18
4t = 18
t = {{{18/4}}}
t = 4.5 hrs for the 2nd cyclist to catch the 1st
:
:
Check solution (1st cyclist travel time = 7.5 hr)
10 * 4.5 = 45 mi
6 * 7.5 = 45 mi</PRE>