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Math Help - Motion - Please, any help???

  1. #1
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    Exclamation Motion - Please, any help???

    A large stone is falling through a layer of mud and its depth x m below ground level at time t minutes is given by x = 12 - 12e^(-0.5t). How long will it take the stone to reach within 1 mm of its final position?
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  2. #2
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    Quote Originally Posted by Sunyata View Post
    A large stone is falling through a layer of mud and its depth x m below ground level at time t minutes is given by x = 12 - 12e^(-0.5t). How long will it take the stone to reach within 1 mm of its final position?

    x = 12 - 12e^(-0.5t).

    The given problem can be written as
    12 - x = 12e^(-0.5t)
    !2 - x = 10^-3 m.
    So e^(-0.5t) = (10^-3)/12
    Or e^(0.5t) = 12000
    Take ln on both side and solve for t.
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  3. #3
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    Hello, Sunyata!

    A large stone is falling through a layer of mud.
    Its depth x m below ground level at time t minutes is given by: . x \:=\: 12 - 12e^{-\frac{1}{2}t}

    How long will it take the stone to reach within 1 mm of its final position?

    The stone reaches its final position as t \to\infty.

    . . \lim_{t\to\infty}x \;=\;\lim_{t\to\infty}\left(12 - \frac{12}{e^{\frac{1}{2}t}}\right) \;=\;12 - 0 \;=\;12

    The final positon is 12 meters below ground level.



    When will the stone be 11.99 m below ground level?

    We have: . 12 - 12e^{-\frac{1}{2}t} \;=\;11.99 \quad\Rightarrow\quad -12e^{-\frac{1}{2}t} \;=\;-0.01

    . . . . . . e^{-\frac{1}{2}t} \;=\;\frac{0.01}{12} \quad\Rightarrow\quad -\frac{1}{2}t \;=\;\ln\left(\frac{0.01}{12}\right)

    . . . . . . t \;=\;-2\ln\left(\frac{0.01}{12}\right) \;=\;14.18015367


    Answer: about 14.2 minutes.

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