SOLUTION: Two cyclists start biking from a trails start 3 hrs apart. The second cyclist travels at 10 mph and starts 3 hours after the first cyclist who is traveling at 6 mph. How much time
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Question 252056: Two cyclists start biking from a trails start 3 hrs apart. The second cyclist travels at 10 mph and starts 3 hours after the first cyclist who is traveling at 6 mph. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started cycling. I know d=rt. I tried
d=10(t-3)/6t?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
You are close. I appreciate that you mention what you tried.
What will be the same for both of them?
not the speed not the time but the distance.
d=rt
r1*t1=d1
r2*t2=d2
d1=d2
r1*t1=r2*t2=
t1=t+3
t2=t
6*(t+3)=10t
t=9/2
t=4.5
so instead of dividing them we make them equal.
Using t for the 10 mph means that the time t will be the answer to the question asked about the second cyclist. I also don't like dealing with negatives.
check
6*7.5=10*4.5
45=45
ok
BTW
We could have used t-3
6*(t)=10(t-3)
t=15/2=7.5
but we weren't asked about this guy
so we have to subtract 3
7.5-3=t-3
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