SOLUTION: I previously asked a question regarding chasing someone with a 10 mile head start and I was running at a rate that is two miles per hour faster than the person I was chasing also w

Question 154031: I previously asked a question regarding chasing someone with a 10 mile head start and I was running at a rate that is two miles per hour faster than the person I was chasing also with a hornet flying back and forth between us at a rate of 20 miles per hour. How far would the hornet fly by the time I reached the other person? Stan did a great job of answering the question for me, 5 hous to catch up to the person and the hornet flew 5X20=100 miles. My problem is that I still can not see how to write this out using the formulas. Any help is appreciated. Thank you.
Answer by ptaylor(2198) (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 required to catch the front runner
Front runner runs at the rate of r
You run at the rate of r+2
Distance front runner runs =10+r*t (already has a 10 mi head start)
Distance you run=(r+2)*t=rt+2t
Now when the above two distances are equal, you will have caught the front runner, so:
10+rt=rt+2t
subtract rt from each side
10+rt-rt=rt-rt+2t collect like terms
2t=10
t=5 hours
Distance hornet flew (d)=Rate of hornet (20 mph)* amount of time the hornet flew (5 hr)
d=20*5=100 mi
Just as Stan sez
Hope this helps---ptaylor
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