Question 221380
Two cyclists start biking from a trail'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 overtakes the 1st, they will have traveled the same distance.
Write a distance equation using this fact: 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 from the time the 2nd cyclist starts
;
:
Check solution by finding the distance traveled by each (should be equal)
10*4.5 = 45 mi
6*7.5 = 45 mi; confirms our solution
:
:
Answering your question about the time.
it should be apparent to you that the 1st cyclist was on the road 3 hr longer
than the 2nd cyclist "t",  hence we call 1st cyclist time "t+3"