A tennis ball with a speed of 12.5 m/s is
moving perpendicular to a wall. After striking
the wall, the ball rebounds in the opposite
direction with a speed of 11.4 m/s.
If the ball is in contact with the wall for
0.0139 s, what is the average acceleration of
the ball while it is in contact with the wall?
Take “toward the wall” to be the positive
direction.
Answer in units of m/s2
-1719 m/s^2.
thank you for the help solving the problem
Well, looks like this tennis ball really knew how to make an impact! Let's calculate its average acceleration while it was in contact with the wall.
To get the average acceleration, we need to find the change in velocity and divide it by the time of contact. The initial velocity of the ball was 12.5 m/s, and after hitting the wall, it rebounded with a velocity of 11.4 m/s in the opposite direction. That means the change in velocity is the difference between these two speeds, which is 12.5 m/s - (-11.4 m/s), or 23.9 m/s.
Given that the time of contact is 0.0139 s, we can now calculate the average acceleration. We divide the change in velocity by the time of contact:
Average acceleration = change in velocity / time of contact
= 23.9 m/s / 0.0139 s
And the grand finale is...
The average acceleration of this tennis ball during its wall encounter was approximately 1,717.99 m/s²! That's quite a speed workout!
Hope that explanation bounced right into your funny bone!
To find the average acceleration of the ball while it is in contact with the wall, we can use the equation:
acceleration = (change in velocity) / (time)
The change in velocity can be calculated by subtracting the initial velocity of the ball before hitting the wall from the final velocity of the ball after rebounding. In this case, the initial velocity is 12.5 m/s (towards the wall), and the final velocity is -11.4 m/s (opposite direction after rebounding). Since the final velocity is in the opposite direction, we assign it a negative sign.
change in velocity = final velocity - initial velocity
= -11.4 m/s - 12.5 m/s
= -23.9 m/s
The time the ball is in contact with the wall is given as 0.0139 s.
Now we can substitute the values into the equation:
acceleration = change in velocity / time
= -23.9 m/s / 0.0139 s
= -1720.86 m/s²
The negative sign indicates that the acceleration is directed towards the wall (the positive direction defined in the problem statement).
Therefore, the average acceleration of the ball while it is in contact with the wall is -1720.86 m/s².
a=(V-Vo)/t
a=(-11.4-12.5)/0.0139 = -1719 m/s^2.