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Math Help - [SOLVED] Find the velocity of the particle at time t

  1. #1
    Junior Member moonman's Avatar
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    [SOLVED] Find the velocity of the particle at time t

    For 4.95 seconds, a particle moves in a straight line according to the position function:
    f(t) = (e^(t)) (5 - t) -5

    Find the velocity of the particle at t = 0, t = 1.

    Okay the book states "the velocity function is the derivative of the position function. So I'm assuming I have to take the derivative of the function f(t)?
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by moonman View Post
    For 4.95 seconds, a particle moves in a straight line according to the position function:
    f(t) = (e^(t)) (5 - t) -5

    Find the velocity of the particle at t = 0, t = 1.

    Okay the book states "the velocity function is the derivative of the position function. So I'm assuming I have to take the derivative of the function f(t)?
    You got it. Then just evaluate f'(0) & f'(1)
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  3. #3
    Junior Member moonman's Avatar
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    Thanks for help,
    So for the derivative of f(t) I got -5 is that correct?
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  4. #4
    Member sinewave85's Avatar
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    Quote Originally Posted by moonman View Post
    For 4.95 seconds, a particle moves in a straight line according to the position function:
    f(t) = (e^(t)) (5 - t) -5

    Find the velocity of the particle at t = 0, t = 1.

    Okay the book states "the velocity function is the derivative of the position function. So I'm assuming I have to take the derivative of the function f(t)?
    Right! So:

    f^{\prime}(t) = (e^{t})(5 - t)^{\prime} + (e^{t})^{\prime}(5 -t) + (-5)^{\prime}

    f^{\prime}(t) = e^{t}(-1) + e^{t}(5 - t)

    f^{\prime}(t) = e^{t}(4 - t)

    and then

    f^{\prime}(0) = e^{0}(4 - 0) = 4

    f^{\prime}(1) = e^{1}(4 - 1) = 3e \approx 8.1548
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  5. #5
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by moonman View Post
    Thanks for help,
    So for the derivative of f(t) I got -5 is that correct?
    Nope that is not quite right. You need to use the product rule for your derivative. Look at sinewave's post.
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  6. #6
    Junior Member moonman's Avatar
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    Now find when the particle is at rest.

    Now I need to find when the particle is at rest.
    Okay, so the particle will be at rest when the velocity is zero right? v(t)=0

    The velocity function I have is (e^(t)) * (4 - t)

    So if I set this up like this:

    (e^(t)) * (4 - t) = 0

    and then solve for "t" I should get when the particle is resting?
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  7. #7
    Junior Member moonman's Avatar
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    So when t = 0 and when t = 4 ? Right?
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  8. #8
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by moonman View Post
    So when t = 0 and when t = 4 ? Right?
    Not exactly
    at t= 0

    e^{t} = e^{0} = 1

    at t=0
    4-t = 4
    --------------------
    So at t=0

    (e^(t)) * (4 - t) = 1*4 = 4

    _____________________________________________

    But when t= 4 , 4-t =0
    Hence

    (e^(t)) * (4 - t) = 0 = (e^ 4 ) * 0 = 0


    Hence t=4 is correct & not t= 0
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