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Math Help - Time-Distance Issue

  1. #1
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    Time-Distance Issue

    Hello , I am having trouble with the following question:
    Tom and bill agreed to race across 50-foot pool and back again.They started together but Tom finished 10feet ahead of bill.IF there rates were constant and Tom finished the race in 27 seconds , how did it take bill to finish:
    The answer is supposed to be 30 but i am getting 33.75

    Here is how i am doing it:
    Toms velocity is: 50/27 ft/sec
    Bills velocity will be : 40/27 (Because Tom finished 10ft ahead of Bill)
    So Time taken for bill to finish the race will be:
    S/V = 50 / (50/27) = 33.75
    Could anyone tell me what i am missing or doing wrong!! Thanks
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  2. #2
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    Re: Time-Distance Issue

    Tom and bill agreed to race across 50-foot pool and back again.
    Tom's avg speed ... \frac{100 \, ft}{27 \, sec}

    Bill's ... \frac{90 \, ft}{27 \, sec}
    Thanks from MikeNoob
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  3. #3
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    Re: Time-Distance Issue

    Tom and bill agreed to race across 50-foot pool and back again. 100 instead of 50.
    Thanks from MikeNoob
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  4. #4
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    Re: Time-Distance Issue

    Now i get it. The time is for the two sides... Thanks.
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