<PRE><B>Question 204814</B>
Two cyclists start biking from a trails 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 caches up with the first from the time the second cyclist started biking?
:
Let t = travel time of the 2nd cylist
then
(t+3) = travel time of the 1st cylist
:
When the 2nd catches the 1st, they will have traveled the same distance
Write dist 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 to catch the 1st
:
:
Check solution by finding the distance of each (should be equal)
10 * 4.5 = 45 mi
6 * 7.5 = 45 mi</PRE>