SOLUTION: 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
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Question 204814: 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? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 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 =
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