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Thread: Does not compute!

  1. #1
    Junior Member
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    Does not compute!

    A word problem is given which I've answered correctly, but a second question doesn't add up. The problem is:

    A North-South highway intersects an East-West highway at a point P. An automobile crosses P at 10:00 AM, traveling east at a constant rate of 20 mph. At that same instant another automobile is two miles north of P, traveling south at 50 mph. Find a formula which expresses the distance d between the automobiles at time t (hours) after 10:00 AM. At what time will the automobiles be 104 miles apart?

    My answer, which the book confirms, is

    $\displaystyle d = \sqrt{2900t^2 - 100t + 4}$

    To answer the second question, I started with

    $\displaystyle 104 = \sqrt{2900t^2 - 100t + 4}$

    $\displaystyle \Rightarrow 10816 = 2900t^2 -100t + 4$

    $\displaystyle \Rightarrow 2704 = 725t^2 - 25t + 1$

    $\displaystyle \Rightarrow 725t^2 - 25t - 2703 = 0$

    Using the Quadratic Formula, I get

    $\displaystyle \frac {25 \pm \sqrt {625 + 7838700}} {1450}$

    which I can only reduce to

    $\displaystyle \frac {5 \pm \sqrt {313573}} {290}$

    But the answer given in the book is:

    $\displaystyle \frac {1 + \sqrt 30} {29}$

    So how do I get there? Did I make a mistake along the way?
    Last edited by earachefl; Jul 16th 2007 at 06:19 AM. Reason: mistake
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  2. #2
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    Hello,earachefl!

    A North-South highway intersects an East-West highway at a point $\displaystyle P$.
    An automobile crosses $\displaystyle P$ at 10:00 AM, traveling east at a constant rate of 20 mph.
    At that same instant another automobile is two miles north of $\displaystyle P$, traveling south at 50 mph.
    (a) Find a formula which expresses the distance $\displaystyle d$ between the automobiles
    . . .at time $\displaystyle t$ (hours) after 10:00 AM.
    (b) At what time will the automobiles be 104 miles apart?

    I don't agree with either of their answers . . .
    Code:
              Q *
                |
            50t |
                |
              B *
                |  *   d
          2-50t |     *
                |        *
          - - P * - - - - - * - -
                |    20t    A
                |

    At 10 AM, the first car is at point $\displaystyle P$.
    . . $\displaystyle t$ hours later, it is at point $\displaystyle A\!:\;PA = 20t$

    At 10 AM, the second car is at point $\displaystyle Q\!:\;QP = 2$.
    . . $\displaystyle t$ hours later, it is at point $\displaystyle B\!:\;QB = 50t$.
    . . Hence: .$\displaystyle BP = 2-50t$

    Pythagorus says: .$\displaystyle d^2\:=\:(20t)^2 + (2-50t)^2$

    (a) Then we have: .$\displaystyle \boxed{d \:=\:\sqrt{2900t^2 - {\color{red}200}t + 4}}$


    If $\displaystyle d = 104\!:\;\;\sqrt{2900t^2 - 200t + 4} \:=\:104$

    Square both sides: .$\displaystyle 2900t^2 - 200t + 4 \:=\:10816\quad\Rightarrow\quad 2900t^2 - 200t - 10812 \:=\:0$

    Divide by 4: .$\displaystyle 725t^2 - 50t - 2703 \:=\:0$

    Quadratic Formula: .$\displaystyle t \;=\;\frac{-(-50) \pm\sqrt{(-50)^2 - 4(725)(-2703)}}{2(725)} $

    . . $\displaystyle t \;=\;\frac{50\pm\sqrt{7841200}}{1450} \;=\;\frac{50 \pm 20\sqrt{19603}}{1450} \;=\;\frac{5 \pm 2\sqrt{19603}}{145} $


    And the positive root is: .$\displaystyle \boxed{t \:=\:\frac{5 + 2\sqrt{19603}}{145} \:\approx\:1.966\text{ hours.}}$

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