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Math Help - 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).

    D(t)

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

    <br />
D_A(t)=66t

    <br />
D_B(t)=112t

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

    So, we have that:

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

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


    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 t hours?
    (c) When are they 1300 miles apart?
    We can solve all the problems at once . . .
    Code:
        P *
          |  *
          |     *   d
      66t |        *
          |           *
          |              *
          * - - - - - - - - *
          O      112t

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

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

    Their distance d is the hypotenuse of right triangle POQ.

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


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

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

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

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