physics
posted by renee
At a certain time a particle had a speed of 91 m/s in the positive x direction, and 4.7 s later its speed was 73 m/s in the opposite direction. What was the average acceleration of the particle during this 4.7 s interval?

Henry
a = (Vf  Vo) / t.
a = (73  91) / 4.7 = 34.9m/s^2.
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