SOLUTION: 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
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Question 128131: 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?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let t=time that will pass before the second cyclist catches up with the first from the time the second cyclist started biking
Now we know that when they have both travelled the same distance, the second cyclist will have overtaken the first
Distance first cyclist travels=6t+6*3=6t+18 (note: the first cyclist has travelled 6*3 or 18 miles before the second cyclists starts)
Distance second cyclist travels=10t
Now we know that these two distances must be equal, so:
6t+18=10t subtract 6t from both sides
6t-6t+18=10t-6t collect like terms
18=4t divide both sides by 4
t=4.5 hrs
Ck
6*4.5+18=10*4.5
27+18=45
45=45
Hope this helps---ptaylor
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