Question

At a certain time a particle had a speed of 16 m/s in the positive x direction, and 2.2 s later its speed was 30 m/s in the opposite direction. What is the magnitude of the average acceleration of the particle during this 2.2 s interval?

Answer 1

change in velocity = end - begin = -30 - 16 = -46 m/s

a = delts v/ delta t = -46/2.2 = -20.9 m/s^2

Answer 2

Thank you so much!