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Thread: Ahh please help!

  1. #1
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    Ahh please help!

    Car A and Car B leave the origin at noon. Car A travels north at 66 mph and car B travels east at 112 mph.
    (a) How far apart are they after 1 hour?
    (b) How far apart are they after hours?
    (c) When are they 1300 miles apart?

    I know (a) is 130 miles. But I can't seem to get b, and therefore i can't do c. If anyone can help I'd greatly appreciate it =)
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  2. #2
    Senior Member topher0805's Avatar
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    What you are looking for is distance (D) as a function of time (t).

    $\displaystyle D(t)$

    Well, we know the equations that give us the distance of Car A and Car B from the origin:

    $\displaystyle
    D_A(t)=66t$

    $\displaystyle
    D_B(t)=112t$

    And we know that to get the distance between the two we use Pythagoras Theorem.

    So, we have that:

    $\displaystyle D_{AB}(t)=\sqrt {(D_A)^2+(D_B)^2}$

    We then add in our knowledge of $\displaystyle D_A$ and $\displaystyle D_B$ to get:


    $\displaystyle D_{AB}(t)=\sqrt {(66t)^2+(112t)^2}$
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  3. #3
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    Thank you so much!
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  4. #4
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    Hello, lilikoipssn!

    Car A and Car B leave the origin at noon.
    Car A travels north at 66 mph and car B travels east at 112 mph.

    (a) How far apart are they after 1 hour?
    (b) How far apart are they after $\displaystyle t$ hours?
    (c) When are they 1300 miles apart?
    We can solve all the problems at once . . .
    Code:
        P *
          |  *
          |     *   d
      66t |        *
          |           *
          |              *
          * - - - - - - - - *
          O      112t

    Car $\displaystyle A$ drives north at 66 mph.
    . . In $\displaystyle t$ hours, it has gone $\displaystyle 66t$ miles to point $\displaystyle P$

    Car $\displaystyle B$ drives east at 112 mph.
    . . In $\displaystyle t$ hours, it has gone $\displaystyle 112t$ miles to point $\displaystyle Q.$

    Their distance $\displaystyle d$ is the hypotenuse of right triangle $\displaystyle POQ.$

    . . $\displaystyle d \;=\;\sqrt{(66t)^2 + (112t)^2} \;=\;\sqrt{16900t^2} \;=\;130t$ miles.


    (a) When $\displaystyle t = 1\!:\;\;d \:=\:130(1) \:=\:130$ miles.

    (b) At time $\displaystyle t$, their distance is: .$\displaystyle d \:=\:130t$ miles.

    (c) If $\displaystyle d = 1300$, we have: .$\displaystyle 130t \:=\:1300 \quad\Rightarrow\quad t \:=\: 10$ hours.

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