# 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?

1. Henry

a = (Vf - Vo) / t.

a = (-73 - 91) / 4.7 = -34.9m/s^2.

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