222k views
4 votes
A tennis ball is moving at a velocity of +54 m/s before a tennis player returns it with a velocity of -58 m/s. If the ball was in contact with her racket for 0.8 s, what was its acceleration? (Assume the acceleration was constant throughout the hit.)

A. -140 m/s2
B. -5 m/s2
C. 72.5 m/s2
D. 112 m/s2

User Ranju R
by
5.3k points

2 Answers

2 votes

Answer:

a. 140 m/s2

Step-by-step explanation:

User Jae Sung Chung
by
5.6k points
6 votes

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning)

Change in speed = (-58 m/s) minus (54 m/s)

Change on speed = -112 m/s

Acceleration = (-112 m/s) / (0.8 s)

Acceleration = -140 m/s²

User Armen Markossyan
by
5.8k points