Velocity as a function of position x

[vela]

You didn't integrate correctly.

 

[nasu]

Ok I have got myself confused I was thinking too much of the problem and did not realize how simple it is:)

All I do from

v*dv/dx= 1/m(Fo+cx)

Is multiply by dx so I have v(x)=(Fo+cx^2)/2m

??

What do you get after just multiplying by dx? Before integration?

 

[KiNGGeexD]

vdv=1/m(Fo+cx) dx ?

 

[KiNGGeexD]

Or would it be

v dx dv=1/m(Fo+cx) dx

 

[nasu]

vdv=1/m(Fo+cx) dx ?

OK. So far so good.

Then what is $\int vdv$?
And what is $\int \frac{1}{m} (F_o+cx) dx$?

 

[KiNGGeexD]

vdv=(Fo+cx)/m dx

So

v^2/2= (Fo+x^2)/2m

So

v(x)= sqrt of the above lol

 

[nasu]

Not right yet.
The integral over v is OK.
The other one is not.
You have a sum there but you integrate only one term of the sum. Kind of. The 1/2 goes only to the last term.

So, how do you integrate a sum of terms?

 

[KiNGGeexD]

(Fox+x^2)/2mx

 

[nasu]

Are you trying random guesses again?
How do you deal with a sum, when you need to integrate? You did not answer this.
Suppose you have to integrate f(x)=a+bx. What is the integral of f(x)?

 

[KiNGGeexD]

I may word this odd but it is that the integral of the sum is the sum of the separate integrals of the separate components?

So (Fo+cx) dx

Is the same as

Fo dx + cx dx

?

 

[nasu]

Yes. This is correct.
Now integrate each piece separately.

 

[KiNGGeexD]

Ok thank you:)!

 

[KiNGGeexD]

So

v=sqrt of Fo*x+cx^2/m

??

 

[nasu]

If you don't use any parentheses, what you wrote reads like this:
$v=\sqrt{F_o \cdot x +\frac{cx^2}{m}}$
Is this what you mean?
If yes, it's not right. But it's getting closer. :)
I wonder why won't you write it step by step rather than throwing shots in the dark?

 

[KiNGGeexD]

It wasn't a shot in the dark I integrated

Fo dx

So I get Fo*x?

Integrate cx and I get cx^2/2

I'm really not sure where I am going wrong?

 

[KiNGGeexD]

Fo*x+cx^2/2m

Is what I get before isolating v

Which is v^2/2...

 

[nasu]

Isn't the Fo term divided by m as well?

 

[KiNGGeexD]

Yea I meant for the whole thing to be divided by m... Sorry it's my phone and the way I have to type isn't ideal sorry if I'm causing any frustration:(

 

[nasu]

Oh, I was afraid that you may get frustrated.
OK, the whole thing is divided by m. But not by 2. No matter how you interpret it, it is not completely right.
So it should be
$\frac{v^2}{2}=\frac{F_o \cdot x +\frac{cx^2}{2}}{m}$
or
$\frac{v^2}{2}=\frac{F_o \cdot x}{m} +\frac{cx^2}{2m}$
Right?
Have you got this, so far?

 

[KiNGGeexD]

Yea haha!

What I wrote before was before I isolated for the v term:)

I will need to get back to you on those formula because it is hard to decipher on my phone:(

 

[KiNGGeexD]

Bit from what I can gather yes I got what you have

 

[nasu]

Yea haha!

What I wrote before was before I isolated for the v term:)

I will need to get back to you on those formula because it is hard to decipher on my phone:(

I don't know what you mean. The v terms was on the left hand side by itself. It looks pretty isolated to me.

If it's hard to use LaTex, you can still use parentheses, to make the equation unambiguous.
The first equation I wrote in the previous post will look like this:
v^2/2=(Fo*x+cx^2/2)/m.
The second one
v^2/2=(Fo*x)/m + (c*x^2)/(2m)

Now you have to do something about the 1/2 in the v term and then extract square root.
And you are done.

 

[KiNGGeexD]

Yea that's what I done but I didn't separate the equation into two separate parts over m;)

 

[KiNGGeexD]

:) didn't mean to do a wink lol

 

[nasu]

No problem. ;)

 

[KiNGGeexD]

Thanks again! You've been great:)