Average Velocity in One Dimension

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SUMMARY

The discussion focuses on the concept of average velocity in one-dimensional motion with constant acceleration, specifically under the influence of gravity. A scenario is presented where a stone is thrown to a height and then lands below the initial height, illustrating that the average velocity over the entire trip can be negative. In this case, with an initial height of h=0 and a landing point at -3 over 3 seconds, the average velocity is calculated as -1 units/second. This negative value indicates a net displacement downward, emphasizing that average velocity is defined as total displacement divided by time, and it is a vector quantity that includes direction.

PREREQUISITES
  • Understanding of vector quantities and their properties
  • Basic knowledge of kinematics and motion equations
  • Familiarity with the concept of displacement in physics
  • Knowledge of instantaneous velocity and its calculation
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  • Study the principles of kinematics in one-dimensional motion
  • Learn about the equations of motion under constant acceleration
  • Explore the concept of instantaneous velocity and its mathematical derivation
  • Investigate the implications of negative velocity in physical contexts
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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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