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  1. #1
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    college mechanics

    A particle of mass 2kg moves in the xy plane under the action of a force F where F = 2i-6j
    initially, the particle is at the point whose position vector is i+j with a velocity vector i-j
    find the position vector of the particle at any time t


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  2. #2
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    in the x-direction ...

    a_x = \frac{F_x}{m} = 1

    since the force remains constant ...

    x = x_0 + v_{0x} t + \frac{1}{2}a_x t^2

    x = 1 + t + \frac{1}{2}t^2

    in the y-direction ...

    a_y = \frac{F_y}{m} = -3

    since the force remains constant ...

    y = y_0 + v_{0y} t + \frac{1}{2}a_y t^2

    y = 1 - t - \frac{3}{2}t^2

    so ...

    \vec{r} = \left(1 + t + \frac{1}{2}t^2\right)\vec{i} + \left(1 - t - \frac{3}{2}t^2\right)\vec{j}
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  3. #3
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    Quote Originally Posted by divine_mail View Post
    A particle of mass 2kg moves in the xy plane under the action of a force F where F = 2i-6j
    initially, the particle is at the point whose position vector is i+j with a velocity vector i-j
    find the position vector of the particle at any time t


    Much Appreciated
    We can do this a bit more directly.
    \vec{a} = \frac{\vec{F}}{m} = \vec{i} - 3 \vec{j}

    So
    \vec{x} = \overrightarrow{x_0} + \overrightarrow{v_0}t + \frac{1}{2} \vec{a}t^2

    giving
    \vec{x} = (\vec{i} + \vec{j} ) + (\vec{i} - \vec{j} ) t + \frac{1}{2} ( \vec{i} - 3 \vec{j} ) t^2
    which is skeeter's answer.

    -Dan
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