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Thread: Total distance Issue

  1. #1
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    Total distance Issue

    Total distance Issue-img_0524.jpg

    A bit stuck on this one.

    The original equation is x = 1/3t^3 etc etc

    Do I set X equal to Y and t is pretty much X?

    And do I enter it in on my graphic calculator as such?

    A bit confused on the t=3 question too. Total distance isnt the point of (t,y) is it?
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  2. #2
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    Re: Total distance Issue

    Hello, minneola24!

    $\displaystyle \text{A particle moves along the }x\text{-axis in such a way that}$
    $\displaystyle \text{its position at time }t\text{ is given by: }\:x(t)\:=\:\tfrac{1}{3}t^3 - 3t^2 + 8t$

    $\displaystyle \text{When }t = 3\text{, what is the total distance the particle has traveled?}$

    Solve $\displaystyle v \,=\,\tfrac{dx}{dt}\,=\,0$

    . . $\displaystyle t^2 - 6t + 8 \:=\:0 \quad\Rightarrow\quad (x-2)(x-4) \:=\:0 \quad\Rightarrow\quad x \:=\:2,\,4$

    Hence, at $\displaystyle t = 2$ and $\displaystyle t=4$, the particle stops and changes direction.


    $\displaystyle \text{At }t = 0\!:\;x(0) \:=\:\tfrac{1}{3}(0^3) - 3(0^2) + 8(0) \:=\:0$

    $\displaystyle \text{At }t = 2\!:\;x(2) \:=\:\tfrac{1}{3}(2^3) - 3(2^2) + 8(2) \:=\:\tfrac{20}{3}$
    . . The particle has moved $\displaystyle 6\tfrac{2}{3}$ units to the right.

    $\displaystyle \text{At }t = 3\!:\;x(3) \:=\:\tfrac{1}{3}(3^3) - 3(3^2) + 8(3) \:=\:6$
    . . The particle has moved $\displaystyle \tfrac{2}{3}$ units to the left.


    The particle has traveled: $\displaystyle 6\tfrac{2}{3} + \tfrac{2}{3} \:=\:7\tfrac{1}{3}$ units.
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