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Math Help - Integration of Zero where d^2z/dt^2 = 0

  1. #1
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    Integration of Zero where d^2z/dt^2 = 0

    Suppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial
    velocity −j +2k and moves under a constant force F = −i +2j. Find
    the position and velocity for all time.

    The Equation of motion ma = F implies that,

    a = −i +2j

    which gives the three scalar equations:

    d^2x/dt^2= −1,

    d^2y/dt^2= 2,

    d^2z/dt^2= 0

    integrating for x yields:

    x = −1/2t^2+1

    integrating for y yields:

    y = t^2− t

    I am unable find z, where:

    d^2z/dt^2 = 0

    I need to integrate this equation twice to obtain:

    z = 2t

    But as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.

    Any help or advice would be greatly appreciated as I am finding this area of Applied math confusing.
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  2. #2
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    Quote Originally Posted by GrahamJamesK View Post
    Suppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial
    velocity ˆ−j +2ˆk and moves under a constant force F = −i +2ˆj. Find
    the position and velocity for all time.

    The Equation of motion ma = F implies that,

    a = −i +2ˆj

    which gives the three scalar equations:

    d^2x/dt^2= −1,

    d^2y/dt^2= 2,

    d^2z/dt^2= 0

    integrating for x yields:

    x = −1/2t^2+1

    integrating for y yields:

    y = t^2− t

    I am unable find z, where:

    d^2z/dt^2 = 0

    I need to integrate this equation twice to obtain:

    z = 2t

    But as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.

    Any help or advice would be greatly appreciated as I am finding this area of Applied math confusing.
    \displaystyle\int 0 \ dt=c\Rightarrow\int c \ dt=ct+n \ \ \ c,n\in\mathbb{R}

    If c = 2 and n = 0, you have your z = 2t
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    \displaystyle\int 0 \ dt=c\Rightarrow\int c \ dt=ct+n \ \ \ c,n\in\mathbb{R}

    If c = 2 and n = 0, you have your z = 2t
    Thank you for your reply. I understand this, but how can I arrive at 2 if I assume that "z=2t" is unknown.
    Last edited by GrahamJamesK; January 28th 2011 at 07:51 PM. Reason: typo
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    Quote Originally Posted by GrahamJamesK View Post
    Thank you for your reply. I understand this, but how can I arrive at 2 if I assume that "z=2t" is unknown.
    Why are you assuming it is 2t? If you know that, you just set them equal.

    ct+n=2t\Rightarrow c=2 \ \ n=0\Rightarrow z=2t
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  5. #5
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    Ok that makes sense, thank you.
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  6. #6
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    From \frac{d^2z}{dt^2}= 0, you get \frac{dz}{dt}= C, a constant. But you are told that the velocity at t= 0 is -j+ 2k so that the z component of velocity is 2: the constant is 2. From \frac{dz}{dt}= 2, integrating again gives z(t)= 2t+ D. Since you wer told that the initial position was (1, 0, 0), z(0)= 2(0)+ D= 0 and so D= 0. z= 2t.
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  7. #7
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    Your explanation was very instructive, I was getting confused by the vector labels and the axis labels. Kudos!
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