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Thread: 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; Mar 23rd 2009 at 05:58 AM.
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  2. #2
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    Hello, struck!

    Point $\displaystyle O$ 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 $\displaystyle A$ and $\displaystyle B$ which are initially approaching $\displaystyle O$
    on different roads in the following cases:

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

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

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

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

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


    In right triangle $\displaystyle AOB\!:\;\;D\:=\:d^2 \:=\:(120-20t)^2 + (80-20t)^2
    $

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

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

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

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

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

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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 $\displaystyle \theta$ with due E, where $\displaystyle \tan \theta = \frac{20}{10} = 2$.

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

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

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

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

    Grandad
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