ball contact with wall?

Question

2) A tennis ball with a speed of 12.6 m/s is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a speed of 10 m/s. If the ball is in contact with the wall for 0.014 s, what is the average acceleration of the ball while it is contact with the wall?
I thought the answer was 1614.29 m/s^2 but it wasn't right. Could somebody please help me!!

Answer 1

No its wrong.

Draw a diagram to help....

This is an introduction to velocity vectors. With vectors you must determine both the magnitude (the number) and the DIRECTION (which should be positive or negative).

Your signs for velocities are different because the ball travels in two different directions

So
(*edited)
V = 12.6 m/s + (-10 m/s)
V = 2.6 m/s

Acceleration = Velocity / (time)
Acceleration = (2.6 m/s) / (0.014 seconds)
Acceleration = 142.86 m/s^2

Do a reality check. This makes sense because the ball loses alot of momentum once striking the wall thus reducing its acceleration.

(*I edited this by only changed my notation. The answer and work remains the same, I mislabeled the velocity as average velocity when it is clear that I never took the average.*)

Answer 2

Average acceleration would be defined as change in velocity over change in time. Your initial velocity is 12.6 m/s toward the wall (I will call the positive direction) and your final velocity is 10 m/s away from the wall (negative direction). This would mean that your change in velocity is:

-10 m/s - 12.6 m/s = -22.6 m/s

This change in velocity occurs during a time period of 0.014 sec. so:

average acceleration = change in velocity / change in time

a = -22.6 m/s / 0.014 s = -1610 m/s^2

I would check which way is defined as the positive direction and which way is the negative direction as acceleration is a vector (direction matters).

Note:

Average velocity CANNOT be used as a way to determine average acceleration. Check the definition of average acceleration.

Answer 3

uh, that answer is right.

deltaV is 22.6m/s, deltaT is 0.014 seconds, so delta A is deltaV over DeltaT, which is 1614.285.....etc