Solve 2-D Motion: Find Direction at t=0 & t=3.5

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To determine the direction of motion for the particle at t=0 and t=3.5, the trajectory equations x = (1/2t^3 - 2t^2) m and y = (1/2t^2 - 2t) m must be differentiated with respect to time to find the velocity components. The velocity in the x-direction (vx) and y-direction (vy) can be obtained by differentiating the x and y equations separately. The direction of motion can then be calculated using the arctangent of the ratio of vy to vx, rather than using the position coordinates directly. The incorrect answers likely stemmed from using position values instead of velocity to determine the direction. Properly applying these steps will yield the correct angles for the particle's motion.
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A particle's trajectory is defined as x = (1/2t^3 - 2t^2) m and y = (1/2t^2 - 2t)m where t is in s.

What is the particle's direction of motion, measured from the x axis, at t=0 and t= 3.5, measured in degrees counterclockwise from the x axis.

I started out by plugging in t = 0 and t = 3.5 to the equations given. I then attemped to take arctan(y/x) in both cases to find the direction of motion. my answers for these, 0 degrees and 15.9 degrees respectively, both came back as wrong.

Any clues as to what I am doing wrong/how to do it right?
 
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If you used the given equations, then you have found the LOCATION of the particle. What can you do to get information about the velocity of the particle?
 
Try differentiaing wrt t (for both x and y separately) to find the general equations for velocity.
 
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