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Math Help - Mechanics with vectors question!

  1. #1
    Junior Member
    Joined
    Oct 2009
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    Mechanics with vectors question!

    Alright all ive been attempting this question and ive become stuck.

    This is the question.



    So for the first part i wrote F=Mx^.. (X double dot).
    I also stated that if there was no force acting on the particle then the particle would stay at a constant speed V if we ignored instances such as the particle being on a slope.

    For the second part i wrote (Im using X to denote x hat (x^^), same applied with y.)

    F_x X + F_y Y = m x^.. X + m y^.. Y

    => x^.. = 2 , y^.. = 3x^2

    Now for the third part when writing down the inital conditions this is where i become confused. Shall i write it in plain english such as;

    At time equal to 0, this displacement of the particle x^.. will be 0.

    Im not too sure how to answer that part, or the last part.

    Any help is greatly appricated cheers people
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
    Posts
    1,458
    Hi

    Integrating \frac{d^2x}{dt^2} = 2 leads to \frac{dx}{dt} = 2t + C

    The particle starts at rest at time t=0. This means that \frac{dx}{dt} (t=0) = 0 and therefore C=0 and \frac{dx}{dt} = 2t

    Integrating \frac{dx}{dt} = 2t leads to x = t^2 + K

    The particle starts at the origin at t=0. This means that x(t=0) = 0 and therefore K=0 and x = t^2

    Now substitute x = t^2 into \frac{d^2y}{dt^2} = 3x^2 and solve the same way
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