Physics-Trajectory

A particle's trajectory is described by x =(12t3−2t2)m and y =(12t2−2t)m, where t is in s.

What is the particle's speed at t=0s ?

What is the particle's speed at t=5.0s ?
Express your answer using two significant figures.

What is the particle's direction of motion, measured as an angle from the x-axis, at t=0 s ?
Express your answer using two significant figures.

What is the particle's direction of motion, measured as an angle from the x-axis, at t=5.0s ?
Express your answer using two significant figures.

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  1. A. Speed = 0.

    B. X = 12*5^3 - 2*5^2 =
    Y = 12*5^2 - 2*5 =
    Speed = sqrt(X^2 + Y^2) =

    C.

    D. Tan A = Y/X.
    A = ?

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  2. Henry’s post is incorrect. Instinctively, it might be easier to just plug in the values for t, but in order to find speed from a position equation, you must take the derivative.

    V(x) = x ‘(t)
    V(y) = y ‘(t)

    Then take the magnitude of both components. It may help to draw this on a graph, perhaps a position-time graph and / or velocity-time graph.

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  3. Plug it in

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  4. A. v=2m/s
    C. -90

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  5. B. 17.75
    Because

    sqrt([3/ 2 t^2 - 4t]^2 + [t - 2]^2).

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  6. x =(1/2t^3−2t^2)m and y =(1/2t^2−2t)m

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