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Math Help - distance problem

  1. #1
    Senior Member
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    distance problem

    heres a pic of whats going on: http://img405.imageshack.us/img405/415/untitledyp4.jpg

    Ship A travels south at a speed of 8 miles/hr,

    Ship B travels east at a speed of 16 miles/hr.

    The pic shows both ships at 7am.

    with matlab, plot the distance between the ships as a function of time for the next 4 hours.

    the horizontal axis will show the time of day starting at 7.

    and the vertical will show the distance.


    'If visiblity is at 8 miles, estimate the time when the people on both ships can see each other.'

    ..this may be a d=r*t problem.
    dunno, how to set it up correctly.

    thanks for any help.
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  2. #2
    Super Member

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    Hello, rcmango!

    A challenging problem . . .


    Ship A travels south at a speed of 8 mph.
    Ship B travels east at a speed of 16 mph.
    The pic shows both ships at 7am.

    With matlab, plot the distance between the ships as a function of time for the next 4 hours.
    The horizontal axis will show the time of day starting at 7.
    and the vertical will show the distance.

    If visiblity is at 8 miles, estimate the time when people on both ships can see each other.
    Code:
                                  P
                                  *
                                  |
                                  |8t
                                  |
                                  * A
                                * |
                           d  *   |
                            *     |14-8t
                          *       |
                        *         |
          * - - - - - * - - - - - * C
          Q    16t    B   25-16t

    Ship A starts at P and sails south at 8 mph.
    In t hours, it has sailed 8t miles to point A.
    Note that: .AC .= .14 - 8t

    Ship B starts at Q and sails east at 16 mph.
    In t hours, it has sailed 16t miles to point B.
    Note that: .BC .= .25 - 16t

    Using Pythagorus, the distance between them is:
    . . . . . . . . . . . . ___________________
    . . d .= .AB .= .√(25 - 16t) + (14 - 8t) . ← (graph this)

    . . . . . . . . . . . . . . . . . . . . .___________________
    If the distance is 8 miles: . √(25 - 16t) + (14 - 8t) .= .8

    Square both sides: .625 - 800t + 256t + 196 - 224t + 64t .= .8

    . . which simplifies to the quadratic: .320t - 1024t + 757 .= .0

    . . . . . . . . . . . . . . . . . . . . . . . . _________________
    . . . . . . . . . . . . . . . . . .1024 √1024 - 4(320)(757)
    Quadratic Formula: . t .= .-----------------------------
    . . . . . . . . . . . . . . . . . . . . . . . . 2(320)

    . . and we get two roots: .t .= .1.16, 2.04


    We discard the larger answer.
    . . In 2 hours, both ships will be beyond point C.

    Hence, they will be 8 miles apart in 1.16 hours . . . at about 8:10 am.

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