What is the average velocity equation?

AI Thread Summary
The discussion revolves around the distinction between velocity and speed, emphasizing that context is crucial for understanding the variable 'v' in equations. A specific graph is referenced to determine the average velocity, with one participant questioning the book's assertion that the average velocity is zero. It is clarified that the formula used for calculating average velocity applies only to uniformly accelerated motion, which does not fit the graph's scenario. The participants conclude that the average velocity shown in the graph is not zero, highlighting the importance of correctly interpreting motion types. Understanding the context and conditions of motion is essential for accurate calculations in physics.
Helicobacter
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1. When I see the variable v at some place, how do I know as to whether v is being defined as velocity or speed? On Wikipedia both, velocity and speed, are referring to a formula with v.
2. What is the average velocity of the following: http://img226.imageshack.us/img226/3798/veloc0zt.jpg
(consider one unit=1s resp. 1m/s and the vector of the velocity is always 0°)? According to my book it must be zero: v[average]=0.5(v[final]+v[initial]). Is the book right?
 
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Helicobacter said:
1. When I see the variable v at some place, how do I know as to whether v is being defined as velocity or speed? On Wikipedia both, velocity and speed, are referring to a formula with v.
The context of the problem will tell you which is meant.

2. What is the average velocity of the following: http://img226.imageshack.us/img226/3798/veloc0zt.jpg
(consider one unit=1s resp. 1m/s and the vector of the velocity is always 0°)? According to my book it must be zero: v[average]=0.5(v[final]+v[initial]). Is the book right?
If I'm interpreting that graph correctly, the average velocity (or speed) is certainly not zero! (That formula only makes sense for uniformly accelerated motion, but the motion shown in the graph is not uniformly accelerated: the first half is speeding up, the second half is slowing down.) Did the book really say that the average speed shown in that graph is zero? Or are you applying the formula yourself?
 
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I formed the graph without considering uniformly accelerated motion. I understand now - thank you for your feedback.
 
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