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Math Help - Quadratics

  1. #1
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    Quadratics

    Can you please help me with this question

    Point O is the intersection of 2 roads which cross at right angles; one road runs from north to south, the other from east to west. Car A is 100 m due west of O and travelling east at a speed of 20m/s, and Car B is 80 m due north of O and travelling south at 20 m/s.

    a) show that after t seconds their distance apart, d metres is given by
    d^2 = (100-20t)^2 + (80-20t)^2

    b) show that this simpifies to

    d^2 = 400((5-t)^2 + (4-t)^2 ).

    c)show that the min distance apart of the 2 cars is 10 squareroot 2 m
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  2. #2
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    Re: Quadratics

    At any point in time, car 1 is positioned at \displaystyle \begin{align*} \left( -100 + 20t, 0 \right) \end{align*} and car 2 is positioned at \displaystyle \begin{align*} \left( 0, 80 - 20t \right) \end{align*}.

    If you were to draw a segment between these two points and create a right-angle-triangle, you can evaluate the distance between the two points using Pythagoras.
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    Re: Quadratics

    How can i get (5-t)^2 + (4-t)^2
    from 2t^2 -18t +41
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    Re: Quadratics

    You don't. What you should be doing is taking out common factors BEFORE trying to expand.
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    Re: Quadratics

    what I got was
    d^2 = 800t^2 - 7200t +16400
    I took 400 as the common factor and got
    d^2 = 400(2t^2 -18t +41)

    Was what I did wrong ??
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  6. #6
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    Re: Quadratics

    Mathematically it all appears correct. But your APPROACH is wrong.

    Surely you can see that \displaystyle \begin{align*} (100 - 20t)^2 + (80 - 20t)^2 = [20(5 - t)]^2 + [20(4 - t)]^2 = 400(5-t)^2 + 400(4-t)^2 = 400[(5-t)^2 + (4-t)^2] \end{align*}. When you are trying to get to a factorised form, the best approach is to factorise...
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    Re: Quadratics

    Quote Originally Posted by Prove It View Post
    Mathematically it all appears correct. But your APPROACH is wrong.

    Surely you can see that \displaystyle \begin{align*} (100 - 20t)^2 + (80 - 20t)^2 = [20(5 - t)]^2 + [20(4 - t)]^2 = 400(5-t)^2 + 400(4-t)^2 = 400[(5-t)^2 + (4-t)^2] \end{align*}. When you are trying to get to a factorised form, the best approach is to factorise...


    oh!
    okay
    thnks
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    Re: Quadratics

    in the last part this is what i did:

    d^2 =400(2t^2 +41 -81t)
    d^2 = 800(t-9/2)^2 + 200
    d^2 = 200
    d= 10 squareroot 2

    Is my approach correct or is there a faster way of solving it than completing the square??
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  9. #9
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    Re: Quadratics

    Yes your approach is correct. Well done.
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  10. #10
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    Re: Quadratics

    Quote Originally Posted by prove it View Post
    yes your approach is correct. Well done.
    thanks !! ^__^
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