Question

At a certain time a particle had a speed of 18 m/s in the positive x direction, and 2.4 s later its speed was 30 m/s in the opposite direction.What is the average acceleration of the particle during this 2.4 s interval?

Answer 1

$$a=\frac{v_f-v_i}{t}\\a=\frac{-30-18}{2.4}=-20\frac{m}{s^2}$$