SOLUTION: Two cyclist start biking from a trail's start 3 hours apart, The second cyclist travels @ 10mph and starts 3 hourss after the first cyclist who is traveling @ 6mph. How much time w
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Question 252492: Two cyclist start biking from a trail's start 3 hours apart, The second cyclist travels @ 10mph and starts 3 hourss after the first cyclist who is traveling @ 6mph. How much time will pass before the second cyclist catches up with the first from the time the second cyclist srtarted biking? Found 2 solutions by ankor@dixie-net.com, stanbon:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Two cyclist start biking from a trail's start 3 hours apart,
The second cyclist travels @ 10mph and starts 3 hours after the first cyclist who is traveling @ 6mph.
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 catches the 1st, they will have traveled the same distance
:
Write a distance 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 cyclist to catch the 1st
:
:
Check solution (1st cyclist travel time = 7.5 hr)
10 * 4.5 = 45 mi
6 * 7.5 = 45 mi
You can put this solution on YOUR website! The second cyclist travels @ 10mph and starts 3 hourss after the first cyclist who is traveling @ 6mph. How much time will pass before the second cyclist catches up with the first from the time the second cyclist srtarted biking?
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2nd Cyclist DATA:
rate = 10 mph ; time = x-3 hrs ; distance = rt = 10(x-3) miles
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1st Cyclist DATA:
rate = 6 mph ; time = x hrs ; distance = rt = 6x miles
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Equation:
distance = distance
10x-30 = 6x
4x = 30
x = 7.5 hrs (1st cyclist time)
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x-3 = 4.5 hrs (2nd cyclist time)
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Cheers,
Stan H.
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