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Thread: Cars

  1. #1
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    Cars

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

    HINT GIVEN:

    At t = 0, the cars are 2 miles south and 3 miles east of the intersection, respectively.
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  2. #2
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    Hello, blueridge!

    Did you make a sketch?


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

    The first car starts at $\displaystyle P$, 2 miles south of $\displaystyle X\!:\;\;PX=2$
    In $\displaystyle t$ hours, it has moved $\displaystyle 30t$ miles north to point $\displaystyle A.$
    . . Hence: .$\displaystyle AX \:=\:2-30t$

    The second car starts at $\displaystyle Q$, 3 mile east of $\displaystyle X\!:\;QX = 3$
    In $\displaystyle t$ hours, it has moved $\displaystyle 40t$ miles west to point $\displaystyle B.$
    . . Hence: .$\displaystyle BX \:=\:3-40t$

    From right triangle $\displaystyle AXB$, we have: .$\displaystyle d^2 \:=\:AX^2 + BX^2$

    So we have: .$\displaystyle d \;=\;\sqrt{(2-30t)^2 + (3-40t)^2} $

    . . which simplifies to: .$\displaystyle d \;=\;\sqrt{2500t^2 - 360t +13}$

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