V^2 = vi^2 + 2ad, is v velocity or speed?

[SammyS]

No I did read it and I quoted from the link that contradicts it

Where's the contradiction ?

 

[flyingpig]

v\; is the speed of the center of mass of the body.

 

[SammyS]

What does that contradict?

Stating something is a way that's different than the way it's stated somewhere else, doesn't necessarily mean that there is a contradiction.

I may re-read this whole thread again. If I think I can actually find what you're objecting to, I may attempt to clear it up for you.

 

[flyingpig]

No it clearly states that v is speed.

 

[SammyS]

"it" ?? which it?

 

[SammyS]

No it clearly states that v is speed.

OK!

What IS the definition of speed? - as related to velocity.

 

[flyingpig]

Speed is the magnitude of velocity

 

[SammyS]

For a general velocity vector:

\vec{v}=v_x\,\hat{i}+v_y\,\hat{j}+v_z\,\hat{k}

Therefore:

\vec{v}\cdot\vec{v}={v_x}^2+{v_y}^2+{v_z}^2

The speed is:

v=|\vec{v}|=\sqrt{{v_x}^2+{v_y}^2+{v_z}^2}

So, speed squared is:

v^2={v_x}^2+{v_y}^2+{v_z}^2

It's generally accepted notation that for vector, \vec{A}\,,, that A² can mean either "the square of the magnitude of A" or "the dot product of vector A with itself". The result is the same.

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Now, for the title question: "v^2 = vi^2 + 2ad, is v velocity or speed?":

That kinematic equation is generally used for one dimensional motion. In one-dimensional motion, the direction of a vector is indicated by a + or - sign.

Therefore, (as Tiny-Tim indicated) the answer is that v in this equation can mean either speed or velocity.

 

[flyingpig]

Sammy, I pulled this proof from my Calculus book. It begs the same question with the same level of ambiguity

http://img851.imageshack.us/img851/3677/83855156.th.png

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It says v = r' which is the velocity it also says at the end that

The quantity \frac{1}{2}m|v(b)|^2, that is, half the mass times the square of the speed

 

[tiny-tim]

hi flyingpig!

It says v = r' which is the velocity it also says at the end that

yes, it says that v is the velocity, and and |v| is the speed

… that's correct! ​

 

[flyingpig]

I am referring to the "quantity"

 

[SammyS]

In that proof they exchange  \vec{r}\,'(t)\cdot\vec{r}\,'(t)\  for  \left|\vec{r}\,'(t)\right|^2\,.

In Serway's physics textbook (I have an old edition of your textbook.), in chapter 7, there is a section on the scalar product. The last line of that section shows that for any vector, A:

\vec{A}\cdot\vec{A}=\left|\vec{A}\right|^2

 

[flyingpig]

But |A| itself is a scalar - speed. They are using it interchangeably

 

[SammyS]

Yes, they are used interchangeably .

So, v^2 is the same thing as \vec{v}\cdot\vec{v} and the same thing as \left|\vec{v}\right|^2.