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Math Help - Kinematics, find x given v.

  1. #1
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    Kinematics, find x given v.

    A particle moves in a line so that the velocity v m/s is given by v = 1 + e^(−2x). Find the position, x m , in terms of time t seconds ( t ≥ 0 ), given that x = 0 when t = 0.

    I've tried various methods but I can't seem to get the answer
    x = (1/2) ln [2e^(2t)-1].

    Could the question be wrong? Any hints for my first step?
    I know that I should be equating dx/dt = 1 + e^(−2x). And then, integrate both sides wrt t?
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  2. #2
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    Is your DE

    \dfrac{dx}{dt}=1+e^{-2x} or

    \dfrac{dx}{dt}=1+e^{-2t}?
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  3. #3
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    \dfrac{dx}{dt}=1+e^{-2x}
    according to the question given...
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  4. #4
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    In that case, the equation is linear (autonomous, even), if you view t = t(x). Try that approach. What does that give you?
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  5. #5
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    So t = integrate the inverse of 1+e^{-2x} wrt x?
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  6. #6
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    If you mean multiplicative inverse, then yes.
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  7. #7
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    Yeaps, that's what I meant
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  8. #8
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    Right. So what do you get?
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  9. #9
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    Should I be trying with u-substitution or integration by parts? Both are giving weird answers.
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  10. #10
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    Take a look here. Click Show steps to see how they got it.
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  11. #11
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    Ok, at least I know that my "weird" answer is RIGHT! Thanks for the link!
    Now, I just need to find x.
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  12. #12
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    Sure. Let me know how it goes.
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