What is the average velocity equation?

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SUMMARY

The discussion centers on the distinction between velocity and speed, particularly in the context of calculating average velocity using the formula v[average]=0.5(v[final]+v[initial]). Participants clarify that this formula is applicable only for uniformly accelerated motion, which is not the case in the provided graph. The average velocity depicted in the graph is not zero, contradicting the book's assertion. The conversation emphasizes the importance of context in interpreting the variable 'v' as either velocity or speed.

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  • Knowledge of uniformly accelerated motion principles
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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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