Question

10. As a ball bounces off the floor, its velocity changes from 3.5+downward to 3.4 m/s upward. Find (a) the total change in the ball'svelocity and (b) the average acceleration if the change happens in 0.1 seconds. Assume "up" to be the positive direction and include the correct algebra sign in your answers to convey direction.​

Answer 1

Answer:

The total change in velocity is 6.9 m/s and the average acceleration is 69 m$s^{-2}$

Explanation:

The initial velocity of ball, u = -3.5 m/s   [ minus sign indicates downward direction]

The final velocity of ball, v = +3.4 m/s   [plus sign indicates upward direction]

Time = 0.1 s

Total change in velocity = final velocity - initial velocity

                                        = 3.4 - (-3.5)

                                        = 3.4 + 3.5

                                        = 6.9 m/s

Therefore, the total change in velocity is 6.9 m/s

Average acceleration = total change in velocity / time

                                    = 6.9 / 0.1

                                    = 69 m$s^{-2}$

Therefore, the average acceleration is 69 m$s^{-2}$

Hope you understand it!!