At time t = 14 s, the velocity of a particle moving in the x-y plane is v = 0.06i + 2.21j m/s. By time t = 14.05 s, its velocity has become -0.08i + 2.10j m/s. Determine the magnitude day of its average acceleration during this interval and the angle 8 made by the average acceleration with the positive x-axis. The angle is measured counterclockwise from the positive x-axis. Answers: dav = 0= i m/s² O

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