A cyclist leaves a park at 18 mph. A second cyclist leaves the same park at 20 mph. What distance will they travel before the faster cyclist overtakes the slower cyclist?

Answer 1

first leaves at say noon at 18 mph

2nd leaves at 4pm at 20 mph

2nd catches up to 1st when d=rt = (r+2)(t-4) = 18t = 20(t-4) = 20t -80

20t-18t = 80

2t = 80

t = 80/2 = 40 hours to catch up after starting 4 hours later, if cycling 2mph faster.

in general

18t = 20(t-x) = 20t -20x

2t = 20x

t = 10x where x = the number of hours later that the 2nd rider started

distance traveled to catch up = d=18t = 18(10x) = 180x km where x=number of hours later that 2nd rider started