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The velocity of a particle moving in the x-y plane is given by (6.12i + 3.24j) m/s at time t = 3.65 s. Its average acceleration during the
next 0.02 s is (4.0i + 6.0j) m/s². Determine the velocity v of the particle at t = 3.67 s and the angle between the average-acceleration
vector and the velocity vector at t = 3.67 s.
Answers:
V = ( i
8= i
it i
j) m/s

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