Deriving constant acceleration formula

[Femme_physics]

Since
Ac = Dv/Dt (Ac being constant acceleration)

I can isolate Dv and integrate both sides...did I derived the final equation correctly?http://img685.imageshack.us/img685/211/ssvvssvv.jpg

Is that right?

 

[I like Serena]

Hey!

Your first line is correct.
The second line is not (how did you get it?).
The third line is correct again and would indeed follow from the first line (although with v and v0 instead of vf and vi).

 

[Femme_physics]

Well, I integrated from Vfinal to Vinitial as per the integration formula for "X²" case. Then I integrated acceleration from zero till t. No point in integrating zero because it's zero. Yes point in integrating acceleration and time. That's how I got my equation.

Oh, I made a mistake, it should be AC^2/1+1 multiplied by T^2/1+1

 

[I like Serena]

You appear to have integrated the left hand side to ${v^2 \over 1+1}$.
If you take its derivative with respect to v, you should get what you were integrating, but you won't.What's the derivative of ${v^2 \over 1+1}$ with respect to v?Furthermore, what's the derivative of ${a_c^2 \over 1+1} \cdot {t^2 \over 1+1}$ with respect to t?

 

[Femme_physics]

You appear to have integrated the left hand side to ${v^2 \over 1+1}$.
If you take its derivative with respect to v, you should get what you were integrating, but you won't.

Don't I? Are you sure?..

What's the derivative of ${v^2 \over 1+1}$ with respect to v?

Here->

http://img14.imageshack.us/img14/6018/yepyepm.jpg

Furthermore, what's the derivative of ${a_c^2 \over 1+1} \cdot {t^2 \over 1+1}$ with respect to t?

[/QUOTE]

http://img820.imageshack.us/img820/232/56382450.jpg

 

[I like Serena]

Don't I? Are you sure?..

Here->

http://img94.imageshack.us/img94/3884/derivative.jpg

Almost good.
It should be just "v".
Check your calculation.

Anyway, your integral is ∫dv, which is the same as ∫1.dv.
So you are taking the integral of just "1" with respect to v.
Do you know an expression with "v", that has "1" as its derivative?

http://img820.imageshack.us/img820/232/56382450.jpg

You have the derivative of the t² part right. :)

However, "a" is a constant, meaning it does not change, nor does a² change.
Btw, the end result should match the integral that you are taking, which is just "a".
It is not "a.t".

 

[Femme_physics]

http://img822.imageshack.us/img822/654/wolframj.jpg

 

[I like Serena]

Cheater! ;)

But you didn't do it right.
You should have calculated ∫dv instead of ∫v dv.

And even though wolframalpha did not understand the boundaries v0 and v, you should still have those in your formula.

 

[Femme_physics]

Ohhh...you're right!

http://img545.imageshack.us/img545/76/imacheater.jpg

Oh I'm such a cheater ;) I love this software!

If I weren't caught up with my non-calculus studies I'd have invested in figuring it out on my own. I just wanted to bring this topic to a closure for now and not leave it hanging.
I'll get my calculus straight because I'll need it for a first degree anyway, but all in due time :) Thanks for showing me the software and the way to the solution.

 

[I like Serena]

One thing left: the boundaries v0 and v.

$$∫_{v_0}^v dv = [ v ]_{v_0}^v = v - v_0$$

 

[Femme_physics]

ahh..wasn't sure how to add boundaries in wolfram-- thanks