Question 128131
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