<PRE><B>Question 396381</B>
Two cyclists start biking from a trail&#39;s start 3 hours apart.
 The second cyclist travels at 10 miles per hour and starts 3 hours after the
 first cyclist who is traveling at 6 miles per hour.
 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
:
10t = 6(t+3)
10t = 6t + 18
10t - 6t = 18
4t = 18
t = {{{18/4}}}
t = 4.5 hrs from the time the 2nd cyclist starts
:
:
Check this by finding the actual dist each traveled ( they should be equal)
10 * 4.5 = 45 mi
6 * 7.5 = 45; confirms our solution, t = 4.5 </PRE>