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At t= 0, a particle starts from rest at x= 0, y= 0, and moves in the xy plane with an acceleration >a = (4.0ihat+ 3.0jhat)m/s^2. Assume t is in seconds.
Determine the position of the particle as a function of time t.
Express your answer in terms of the unit vectors ihat and jhat.
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A = 4 i + 3 j
V = integral A dt = 4 t i + 3 t j + initial velocity which is 0
R = integral V dt = 2 t^2 i + (3/2)t^2 j + initial position which is 0posted by Damon
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