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Math Help - Question that has me stumped.

  1. #1
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    Question that has me stumped.

    Carlos leaves Los Angeles on a cross-country car trip at 8:00 AM. He averages 50 Miles per hour.

    Juanita plans to take exactly the same route, but does not leave 9:00 AM. He averages 60 miles per hour.

    Develop a diagram or table to determine at what time Juanita will pass Carlos.

    I've figured out that in between 1 and 2PM the cars will have both traveled 300 miles - but I am finding difficulty in determining when the vehicles will be physically passing.

    Thanks for reading...
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  2. #2
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    how about plotting a graph from the values you have, where they intersect would be where the cars pass, that is if you plot a distance-time graph
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  3. #3
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    Worked like a charm!
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  4. #4
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    Quote Originally Posted by Sarlowbat View Post
    Carlos leaves Los Angeles on a cross-country car trip at 8:00 AM. He averages 50 Miles per hour.

    Juanita plans to take exactly the same route, but does not leave 9:00 AM. He averages 60 miles per hour.

    Develop a diagram or table to determine at what time Juanita will pass Carlos.

    I've figured out that in between 1 and 2PM the cars will have both traveled 300 miles - but I am finding difficulty in determining when the vehicles will be physically passing.

    Thanks for reading...
    Sarlowbat,

    If you need to use a table how about

    Hour Carlos Juanita

    1 50 0 9am
    2 100 60 10
    3 150 120 11
    4 . .
    5 . .
    6 . .

    Also, consider if you let x equal the number of hours it will take for Juanita
    to catch Carlos you would have

    50mph*x number of hours equals 60 mph*x number of hours minus 60 because Juanita stats 1 hour later.
    50x=60x-60
    60=10x
    6=x Six hours or 2pm.
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  5. #5
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    Quote Originally Posted by Sarlowbat View Post
    Carlos leaves Los Angeles on a cross-country car trip at 8:00 AM. He averages 50 Miles per hour.

    Juanita plans to take exactly the same route, but does not leave 9:00 AM. He averages 60 miles per hour.
    You can do this one in your head, Sarlow:

    When J leaves, she's 50 miles behind C (C travelled 50 miles from 8 to 9)

    Since J travels at 10 mph faster than C (60 mph - 50 mph), then it will
    take 50 / 10 = 5 hours to catch up.
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