At time t = 13 s, the velocity of a particle moving in the x-y plane is v=0.05i + 2.54j m/s. By time t = 13.05 s, its velocity has become -0.12i+2.42j m/s. Determine the magnitude day of its average acceleration during this interval and the angle made by the average acceleration with the positive x-axis. The angle is measured counterclockwise from the positive x-axis. Answers: day = 0= i i m/s² O

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At time t = 13 s, the velocity of a particle moving in the x-y plane is v=0.05i + 2.54j m/s. By time t = 13.05 s, its velocity has become
-0.12i+2.42j m/s. Determine the magnitude day of its average acceleration during this interval and the angle made by the average
acceleration with the positive x-axis. The angle is measured counterclockwise from the positive x-axis.
Answers:
dav=
e=
i
i
m/s²
O
Transcribed Image Text:At time t = 13 s, the velocity of a particle moving in the x-y plane is v=0.05i + 2.54j m/s. By time t = 13.05 s, its velocity has become -0.12i+2.42j m/s. Determine the magnitude day of its average acceleration during this interval and the angle made by the average acceleration with the positive x-axis. The angle is measured counterclockwise from the positive x-axis. Answers: dav= e= i i m/s² O
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