Question 310328
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 cyclist catches the 1st, they will have traveled the same distance
Write a 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 travel time for the 2nd cyclist to catch the 1st
:
:
Check solution by finding the distance of each (should be the same)
1st cyclist travel time = 7.5 hrs
6*7.5 = 45 mi
10*4.5 = 45 mi