Average Velocity vs. Instantaneous Velocity

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Homework Help Overview

The discussion revolves around the concepts of average velocity and instantaneous velocity in the context of uniformly accelerated motion, as illustrated by a velocity vs. time graph. Participants explore the relationship between these two types of velocity and how they can be derived from the graph.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between average and instantaneous velocity, referencing the slope of the velocity graph and the implications of constant acceleration. Some question the necessity of deriving formulas, while others attempt to clarify the concepts involved.

Discussion Status

The discussion includes various interpretations of the relationship between average and instantaneous velocity. Some participants provide insights into the graphical representation and the linear nature of the acceleration graph. However, there is no explicit consensus on the necessity of deriving formulas, and the relevance of the original post is questioned.

Contextual Notes

One participant notes that the original post may be outdated, suggesting that the urgency of the inquiry may have diminished over time.

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The graph below shows a plot of velocity vs. time for an object undergoing uniformly accelerated motion. The object has instantaneous velocity v1 at time t1 and instantaneous velocity v2 at time t2. Use the graph and the fact that, when the acceleration is constant, the average velocity can be written as vavg = (v1 + v2)/2 to explain how we know that the instantaneous velocity at the midpoint time is equal to vavg on the interval.

V = Vo + at

X - Xo = volt + .5at2

v2 = vo2 + 2a(X - Xo)

X - Xo = .5(Vo + V)t

Now I know the answer has to do with some how deriving one of these formulas to get it to equal the formula for average veolcity but I can't seem to remember how. I think you some how take the derivative of X - Xo = .5(Vo + V)t to get the answer maybe?
 
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What graph?
 
pretty sure I know what the graph looks like cause I just had this problem in a homework lol.

This answer doesn't require any deriving. Average velocity is simply the slope of the velocity over a certain interval, and instantaneous velocity is the slope at a certain point on the interval. We know that the acceleration graph is simply the graph of the slope of the velocity function.The slope of this graph is clearly linear. And because average velocity and instantaneous velocity are just values of the slope over their intervals or at certain points, if we take the average velocity from t1 to t2 it would be the same as the velocity at the midpoint between t1 and t2.

Hopefully that makes some sense.
 
Last edited:
Stryder_SW said:
pretty sure I know what the graph looks like cause I just had this problem in a homework lol.

This answer doesn't require any deriving. Average velocity is simply the slope of the velocity over a certain interval, and instantaneous velocity is the slope at a certain point on the interval. We know that the acceleration graph is simply the graph of the slope of the velocity function.The slope of this graph is clearly linear. And because average velocity and instantaneous velocity are just values of the slope over their intervals or at certain points, if we take the average velocity from t1 to t2 it would be the same as the velocity at the midpoint between t1 and t2.

Hopefully that makes some sense.

you do realize that s/he posted this over a year ago, and probably doesn't need it anymore
 
mmmm no unfortunately I did not realize that...this was a colossal waste of time.
 

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