The velocity of a particle moving in the x-y plane is given by (6.96i + 4.66j) m/s at time t = 3.70 s. Its average acceleration during the next 0.011 s is (7.5i + 3.2j) m/s². Determine the velocity v of the particle at t = 3.711s and the angle between the average- acceleration vector and the velocity vector at t = 3.711 s. Answers: V = ( i e = i i + i j) m/s
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