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Physics 42 Online
OpenStudy (anonymous):

A particle moves along the x axis. It is initially at the position 0.320 m, moving with velocity 0.130 m/s and acceleration -0.230 m/s2. Suppose it moves with constant acceleration for 6.00 s. (a) Find the position of the particle after this time. (b) Find its velocity at the end of this time interval. Next, assume it moves with simple harmonic motion for 6.00 s and x = 0 is its equilibrium position. (Assume that the velocity and acceleration is the same as in parts (a) and (b).) (c) Find its position. (d) Find its velocity at the end of this time interval.

OpenStudy (anonymous):

I have found part (a), that is the position to be -3.04 m and part (b), that is the velocity to be -1.25 m/s. i need help to answer part (c) and (d). I think i need to set something up as \[x= A \cos (\omega t + \Theta)\]

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