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 mile
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Question 155108: 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?
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=amount of time that passes before the second cyclist catches up with the first from the time the second cyclist starts
distance first cyclist travels=6*3+6*t(3 hour head start)
distance second cyclist travels=10*t
Now when the above two distances are equal, the second cyclist will have caught up with the first cyclist, so our equation to solve is:
10t=18+6t subtract 6t from each side
10t-6t=18-6t
4t=18
t=4.5 hours
CK
3*6+6*4.5=10*4.5
18+27=45
45=45
Hope this helps---ptaylor
