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Math Help - Vector problem

  1. #1
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    Vector problem

    I apologize for the unhelpful title. I don't know how to otherwise classify this problem.

    A particle moves along the graph of y = cos(x) so that its x-component of acceleration is always 2. At time t=0, the particle is at the point ( \pi, -1) and the velocity of the particle is {0,0} (that's the notation my book uses to show a vector).

    (a) find the position vector of the particle.
    Got that. { t^2 + \pi , cos (t^2 + \pi)}

    (b). Find the speed for the particle when it is at the point (4, cos(4))
    I don't get the question.
    Speed is the absolute value of velocity So, it seems like I should set the x part (for lack of a better word) of vector to 4 and the y part of the vector.

    However, the answer is some big complicated thing over a square root.
    Specifically, \sqrt((2t)^2 + 2t(-sin(t^2 +\pi))^2 where t= \sqrt(4 - \pi) .


    Basically, how do I do (b)?

    Thanks!
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  2. #2
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    Quote Originally Posted by Truthbetold View Post
    I apologize for the unhelpful title. I don't know how to otherwise classify this problem.

    A particle moves along the graph of y = cos(x) so that its x-component of acceleration is always 2. At time t=0, the particle is at the point ( \pi, -1) and the velocity of the particle is {0,0} (that's the notation my book uses to show a vector).

    (a) find the position vector of the particle.
    Got that. { t^2 + \pi , cos (t^2 + \pi)}

    (b). Find the speed for the particle when it is at the point (4, cos(4))
    I don't get the question.
    Speed is the absolute value of velocity So, it seems like I should set the x part (for lack of a better word) of vector to 4 and the y part of the vector.

    However, the answer is some big complicated thing over a square root.
    Specifically, \sqrt((2t)^2 + 2t(-sin(t^2 +\pi))^2 where t= \sqrt(4 - \pi) .


    Basically, how do I do (b)?

    Thanks!

    \frac{d^2x}{dt^2} = 2

    \frac{dx}{dt} = 2t + C

    C = 0 since velocity = 0 at t = 0

    \frac{dy}{dt} = -\sin{x} \cdot \frac{dx}{dt} = -2t\sin{x}

    speed = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}

    speed = \sqrt{4t^2 + 4t^2\sin^2{x}}
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