SOLUTION: A Long distance runner started a course running an average speed of 2mph. 1.33 hrs later, a cyclist traveled the same course at an average speed of 4mph. how many hrs after the run

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Question 124356: A Long distance runner started a course running an average speed of 2mph. 1.33 hrs later, a cyclist traveled the same course at an average speed of 4mph. how many hrs after the runner started did the cyclist overtake the runner?
round answer to the nearest tenth.
Answer by ptaylor(2198) About Me  (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 it takes cyclist to overtake runner AFTER the cyclist starts.
And t+1.33 hr=time that we are looking for
Distance runner runs before cyclist starts=2*1.33=2.66 mi
Total distance runner runs=2t+2.66
Total distance cyclist travels=4t
We know that the cyclist will have overtaken the runner whe they have both travelled the same distance. So our equation to solve is:
2t+2.66=4t subtract 2t froom both sides
2t-2t+2.66=4t-2t collect like terms
2t=2.66 divide both sides by 2
t=1.33 hr
t+1.33hr=1.33hr+1.33hr=2.66hr=2.7hr-------------------ans
CK
In 2.7 hr, runner runs2*2.66=5.32 mi
In 1.33hr cyclist travels 4*1.33=5.32mi
Hope this helps--ptaylor