An electronic line judge camera captures the impact of a
57.0-g
tennis ball traveling at
32.8 m/s
with the side line of a tennis court. The ball rebounds with a speed of
21.2 m/s
and is seen to be in contact with the ground for
4.08 ms.
What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground? Assume one-dimensional motion.
The answer is 13235m/s^2 but I don't understand why the unit is in m/s^2 and not just m? Can somebody explain that? I use the equation (Vf-Vi)/T to sole this question.
W
the velocity changes from 32.8m/s downward, to 21.2m/s upward in 4.08ms
acceleration is the rate of velocity change
m/s per sec , or m/s^2
put units in your equation
...(21.2m/s - -32.8m/s) / .00408s
...13200m/s/s ... 13200m/s^2
FYI ... watch your significant figures
The unit for average acceleration is commonly expressed as meters per second squared (m/s^2) because acceleration measures the rate at which velocity changes over time.
Acceleration is defined as the change in velocity divided by the change in time. In this case, the average acceleration of the tennis ball is the change in its velocity (final velocity minus initial velocity) divided by the time it takes for the velocity change to occur.
When you use the equation (Vf - Vi)/T to solve for acceleration, you are essentially finding the change in velocity per unit of time. The numerator (Vf - Vi) represents the change in velocity, while the denominator (T) represents the time interval over which the change occurs.
Since velocity is measured in meters per second (m/s) and time is measured in seconds (s), the result of dividing velocity by time is meters per second per second, or meters per second squared (m/s^2). The unit "per second squared" indicates that the velocity is changing by a certain amount per unit of time squared.
Therefore, in the context of average acceleration, a unit of m/s^2 signifies that the velocity is changing by a certain amount (m/s) for each second (s) that passes. It tells us the rate at which the velocity is changing and indicates the intensity of the acceleration experienced by the tennis ball.
The unit of average acceleration is indeed in m/s^2 (meters per second squared), not just in meters (m). The reason for this is that acceleration measures how quickly an object changes its velocity over time.
In this problem, you are given the initial velocity (Vi) and the final velocity (Vf) of the tennis ball. The time (T) is also given. To calculate the average acceleration, you can use the equation:
average acceleration = (Vf - Vi) / T
The numerator (Vf - Vi) represents the change in velocity of the ball, and the denominator (T) represents the change in time. The units of velocity are in m/s, and the unit of time is in seconds.
When you divide the change in velocity (in m/s) by the change in time (in seconds), you get the average rate at which the velocity of the ball changes over time. This rate is the average acceleration, and its units are m/s^2.
So in this problem, when you plug in the given values:
average acceleration = (21.2 m/s - 32.8 m/s) / (4.08 ms)
You would need to convert milliseconds to seconds before performing the calculation, but once you do, you should get the correct answer of 13235 m/s^2.
Therefore, the unit of average acceleration is in m/s^2 because it represents the change in velocity per unit of time.