SOLUTION: Two runners leave the starting gate, one running 12mph and the other 10mph. If they maintain the pace, how long will it take for them to be 1/4 mile apart?

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Question 156861This question is from textbook
: Two runners leave the starting gate, one running 12mph and the other 10mph. If they maintain the pace, how long will it take for them to be 1/4 mile apart? This question is from textbook

Found 2 solutions by Alan3354, jim_thompson5910:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Two runners leave the starting gate, one running 12mph and the other 10mph. If they maintain the pace, how long will it take for them to be 1/4 mile apart?
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The difference in speed it 2 mph, so they separate at that rate.
1/4 mile at 2 mph will take 1/8 hour, or 7.5 minutes.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let's set up the equation for the faster runner:


d=rt Start with the distance-rate-time formula


d=12t Plug in r=12


So after "t" hours, the faster runner has run 12t miles
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Let's set up the equation for the slower runner:


d=rt Start with the distance-rate-time formula


d=10t Plug in r=10

So after "t" hours, the faster runner has run 10t miles



Now if you want to know the distance between them, simply subtract the two expressions to get


12t-10t


Since we want the distance to be 1%2F4 of a mile, this means that

12t-10t=1%2F4




12t-10t=1%2F4 Start with the given equation.


4%2812t-10t%29=4%281%2Fcross%284%29%29 Multiply both sides by the LCD 4 to clear any fractions.


48t-40t=1 Distribute and multiply.


8t=1 Combine like terms on the left side.


t=%281%29%2F%288%29 Divide both sides by 8 to isolate t.


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Answer:

So the answer is t=1%2F8 which approximates to t=0.125.


So in 1%2F8 of an hour (or 7 and half minutes) the distance between the two runners is 1%2F4 of a mile