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Math Help - Intersection

  1. #1
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    Intersection

    Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 mph. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 mph. Express the distance d between the cars as a function of time t.
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  2. #2
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    Hello, symmetry!

    Two cars are approaching an intersection.
    One is 2 miles south of the intersection and is moving at 30 mph.
    The other car is 3 miles east of the intersection and is moving at 40 mph.
    Express the distance d between the cars as a function of time t.
    Code:
                C     3 - 40t     B   40t   Q
                * - - - - - - - - * - - - - *
                |              *
                |           *
        2 - 30t |        * d
                |     *
                |  *
               A*
                |
            30t |
                |
               P*

    The first car starts at P and drive north to the intersection C.
    In t hours, it has gone 30t miles to point A.
    . . Hence, AC \:=\: 2 - 30t

    The other car starts at Q and drives west to C.
    In t hours, it has gone 40t miles to point B.
    . . Hence, BC \:=\:3 - 40t

    Using Pythagorus, the distance between them is: . d^2\;=\;(2-30t)^2 + (3-40t)^2

    Simplifying, we get: . d^2\;=\;2500t^2 - 360t + 13

    Therefore: . \boxed{d \;=\;\sqrt{2500t^2 - 360t + 13}}

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  3. #3
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    ok

    Like always, your replies are greatly appreciated as I study for my test.
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  4. #4
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    Quote Originally Posted by symmetry View Post
    Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 mph. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 mph. Express the distance d between the cars as a function of time t.
    We have D = r*t

    Distance of the slow car moving at 30 mph is 30t
    Distance of the fast car moving at 40 mph is 40t

    Now, this problem is practically asking you to use the pythag. theorem.

    a^2 + b^2 = c^2

    = sqrt(40t^2 + 30t^2)

    = sqrt(1600t^2 + 900t^2)

    = sqrt(2500t^2)

    And thus,

    = 50*t, which is your final answer.
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