# 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 ?

What is the particle's direction of motion, measured as an angle from the x-axis, at t=0 s ?

What is the particle's direction of motion, measured as an angle from the x-axis, at t=5.0s ?

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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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