Average Velocity in One Dimension

AI Thread Summary
The discussion centers on the average velocity of a stone thrown into the air and its descent under constant acceleration due to gravity. When the stone is thrown to a height and lands lower than its starting point, the average velocity over the entire trip can be calculated as negative, indicating a net displacement in the downward direction. The average velocity is defined as displacement over time, which considers only the initial and final positions, disregarding any "cancelled out" displacement during the ascent. The concept of instantaneous velocity is introduced, explaining that it is derived from calculating mean velocities over increasingly shorter time intervals. Overall, the negative average velocity signifies movement in the opposite direction from the initial throw.
nDever
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A question about the average velocity of bodies undergoing one-dimensional motion and a constant acceleration (gravity in this case).

A case scenario.

Suppose that initially, I throw a stone into the air at a height h. For the sake of argument, let's suppose that even though I threw the stone straight into the air, when it comes back down for the descent, it landed at a point lower than h.

When t=0, the position of the stone is h and at some later time, its position is h again. The average velocity from the initial time to the time when the stone's position is h again is zero because during that time interval, the stone "replaced" all of the distance that it displaced. That I understand. Let's now examine the average velocity from t=0 to the final time when the stone is at some point lower than h. This is where I have questions.

Lets make

h=0,
the stone's landing point is -3,
the entire trip happens over 3 seconds.

So then, the average velocity from [0, 3] is -1 units/second.

Conceptually, what is the meaning of -1 units/second?

Does the calculation disregard the "cancelled out" displacement completely?
 
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nDever said:
So then, the average velocity from [0, 3] is -1 units/second.

Conceptually, what is the meaning of -1 units/second?

Does the calculation disregard the "cancelled out" displacement completely?

Velocity is a vector quantity, meaning that is has a direction as well as a size.

"Ten meters per second" isn't a velocity, it's a speed. Negative speed has no meaning.

"Ten meters per second to the north" is a velocity. "Negative ten meters per second to the north" is the negative of that velocity, and it means that you're traveling to the south.
 
Mean velocity over a time interval is usually defined as displacement over that interval divided by time interval. It is a vector because displacement is a vector. In calculating the displacement, all that counts are the positions of the body at the start and end of the interval.

This definition neatly generates the notion of instantaneous velocity. Suppose we need the instantaneous velocity at time t. We find the mean velocity over a short time interval centred on t. Then we consider shorter and shorter intervals centred on t. The mean velocites so calculated will home in on a limiting value, which is what is meant by the instantaneous velocity at time t.
 
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