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Math Help - Circle and Walking Problem

  1. #1
    Junior Member Godfather's Avatar
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    Circle and Walking Problem

    Thee friends--whose walking rates are 1ft./sec, 3ft/sec, and 6ft./sec--start out together walking in the same direction around a circular track that is 300 feet in circumference. After how many minutes are the three of them together?
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    Member Rimas's Avatar
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    How do u do this??
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Godfather View Post
    Thee friends--whose walking rates are 1ft./sec, 3ft/sec, and 6ft./sec--start out together walking in the same direction around a circular track that is 300 feet in circumference. After how many minutes are the three of them together?
    I can't think of a simple way to do this, perhaps someone else will.

    The difficulty is that once they walk a circumference the position on the circle is "reset."

    Let's start by writing out an equation for the distance each of them has walked in a given time.

    d = vt

    So labelling them "friend 1", "friend 2", and "friend 3" we get that, after a time t each has walked a distance of:
    d1 = (1)t = t
    d2 = (3)t = 3t
    d3 = (6)t = 6t

    Now, we may subtract a circumference from any of these distances if they have walked more than 300 ft. Now, since the speeds are integral multiples of each other we know that they meet when the first friend has finished his/her first circle, so the distance walked is less than or equal to 300 m. (In this time friend 2 has walked the whole circle three times and friend 3 has walked the whole circle 6 times.) So we know that the time required for them to meet up for the first time is less than the time it takes friend 1 to walk the circle:
    300 = t
    or 300 s.

    Let's start this rolling by finding all the times that friends 1 and 2 meet each other.

    We know that friend 2 has to walk around more than once to meet up with friend 1 again, so let's define
    d2 - 300 = d1
    and see when this happens:
    3t - 300 = t

    2t = 300

    t = 150 s.

    When do they meet for a second time?
    d2 - 2*300 = d1

    3t - 600 = t

    2t = 600

    t = 300 s.

    So the second time they meet, they all meet at the starting point. So friends 1 and 2 meet at 150 s and 300 s.

    When do friends 1 and 3 meet for the first time?
    d3 - 300 = d1

    6t - 300 = t

    5t = 300

    t = 60 s

    The second time they meet:
    d3 - 2*300 = d1

    6t - 600 = t

    5t = 600

    t = 120 s.
    etc.

    It is easy to show that friends 1 and 3 meet up every 60 s. Now, friends 1 and 2 meet first at 150 s, which is not a multiple of 60 s, so friends 1, 2, and 3 cannot meet there. But friends 1 and 2 meet again at 300 s, which IS a multiple of 60 s, so they all meet there for the first time.

    So t = 300 s.

    -Dan
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