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Math Help - Word problem using quadratic equations..

  1. #1
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    Word problem using quadratic equations..

    I have to use quadratics to solve this:

    Point O is the intersection of two roads which cross at right angles; one road runs from north to south, the other from east to west. Find the least distance apart of two motorbikes A and B which are initially approaching O on different roads in the following cases:

    (a) A is 120 metres from O travelling at 20 m/s, B is 80 metres from O travelling at 10 m / s.
    Last edited by struck; March 23rd 2009 at 05:58 AM.
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  2. #2
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    Hello, struck!

    Point O is the intersection of two roads which cross at right angles;
    one road runs from north to south, the other from east to west.
    Find the least distance apart of two motorbikes A and B which are initially approaching O
    on different roads in the following cases:

    (a) A is 120 metres from O travelling at 20 m/s.
    . . . B is 80 metres from O travelling at 10 m/s.
    Code:
    -       Q *
    :         |
    :     10t |
    :         |
    80      B *
    :         | *
    :         |   *
    :  80-10t |     * d
    :         |       *
    :         |         *
    :         |           *
    -       O * - - - - - - * - - - - *
                  120-20t   A   20t   P
              : - - - -  120  - - - - :

    Motorbike A starts at P: . OP = 120
    In t seconds, it moves 20t m to point A.
    . . Hence: . OA \:=\:120-20t

    Motorbike B starts at Q: . OQ = 80
    In t seconds, it moves 10t m to point B.
    . . Hence: . OB \:=\:80-10t

    The distance between them is: . d \:=\:AB
    And we wish to minimize d.

    Note that, if we minimize D = d^2, we will minimize d.


    In right triangle AOB\!:\;\;D\:=\:d^2 \:=\:(120-20t)^2 + (80-20t)^2 <br />

    This simplifies to: . D \;=\;500t^2 - 6400t + 20,\!800 .[1]

    The graph of this function is an up-opening parabola.
    . . Its minimum is at its vertex.

    The vertex is at: . t \:=\:\frac{\text{-}b}{2a} \:=\:\frac{\text{-}(\text{-}6400)}{2(500)} \:=\: 6.4

    Substitute into [1]: . D \:=\:d^2 \:=\:500(6.4^2) - 6400(6.4) + 20,\!800 \:=\:320

    Therefore: . d \;=\;\sqrt{320} \;=\;8\sqrt{5} \;\approx\;17.9\text{ m}

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  3. #3
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    Relative velocity: closest approach

    Hello struck
    Quote Originally Posted by struck View Post
    I have to use quadratics to solve this:

    Point O is the intersection of two roads which cross at right angles; one road runs from north to south, the other from east to west. Find the least distance apart of two motorbikes A and B which are initially approaching O on different roads in the following cases:

    (a) A is 120 metres from O travelling at 20 m/s, B is 80 metres from O travelling at 10 m / s.
    I'm not sure why you think this is a Quadratics question. In fact, it's all about relative velocity, and the least distance between two moving bodies.

    In the attached diagram, the Displacement Diagram shows the initial positions of A and B. I've assumed that A is moving due N and B due E.

    The technique is this:

    • Draw a triangle of velocities to find the direction of the velocity of B relative to A.
    • On the displacement diagram, superimpose the direction you've just found to find the displacement of B relative to A.
    • Draw a perpendicular from A onto this line. The length of this perpendicular is the shortest distance between A and B.

    So, in the Triangle of Velocities, you'll see that the velocity of B relative to A is the velocity of B minus the velocity of A.

    This line makes an angle \theta with due E, where \tan \theta = \frac{20}{10} = 2.

    So when this line is drawn on the displacement diagram, the distance OC = 80\tan\theta = 160 \Rightarrow AC = 40.

    So if AP is drawn perpendicular to BC, angle PAC = \theta \Rightarrow AP = 40 \cos\theta.

    You can use a calculator to find \cos\theta, or you can get an exact value (if you know the trig formulae) which is \cos\theta = \frac{1}{\sqrt{5}}

    \Rightarrow AP = \frac{40}{\sqrt{5}}= 8\sqrt{5} m.

    Grandad
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