<PRE><B>Question 315060</B>
The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph.
 How much time will pass before the 2nd cyclist catches up with the 1st from the time the 2nd 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 distance equation; dist = speed * time
2nd dist = 1st 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 first
;
:
Check solution by finding distance of each, should be equal:
10*4.5 = 45 mi
6 *7.5 = 45 mi
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