Integration Newtons Second Law

[Niles]

Homework Statement

Hi

In a book they state
[tex]
\frac{dv}{dt} = a \quad \Rightarrow \quad v(z) = \sqrt{2a(z-z_{max})}
[/tex]
I am trying to reproduce this. Here is what I have so far:

dv/dt = (dv/dz)(dz/dt) = v(dv/dz) = a

Since a is constant (I assume?), I get
[tex]
\int vdv = \int adz \quad \Rightarrow\quad \frac{1}{2}v^2 + C = az
[/tex]
Here C denotes a constant. What should I do from here?Niles.

 

[genericusrnme]

what is z?

if [itex]\frac{dv}{dt} = a[/itex] then [itex]\int dv = \int a\ dt \ \rightarrow v = \int a\ dt[/itex]

 

[tiny-tim]

Hi Niles!

[tex]
\frac{dv}{dt} = a \quad \Rightarrow \quad v(z) = \sqrt{2a(z-z_{max})}
[/tex]
I am trying to reproduce this.
æ
[tex]
\frac{1}{2}v^2 + C = az
[/tex]
Here C denotes a constant. What should I do from here?

erm …

you're there! ​

 

[Niles]

Hi Niles!


erm …

you're there! ​

Thanks. But how is it seen that C=2az_max?

 

[Pengwuino]

Thanks. But how is it seen that C=2az_max?

You really might want to give us a lot more context on what you're doing. What is z? What is [itex]z_{max}[/itex]? If you tell us what the problem is, we can even tell you if the acceleration is constant.

 

[Niles]

It is from a book on how to slow atoms with a Zeeman slower. Here [itex]z[/itex] is the coordinate (longitudinal, along the magnetic field), and [itex]z_{max}[/itex] I believe is the longitudinal coordinate, where the velocity is zero. Sorry, I should have stated that at first.

 

[tiny-tim]

Hi Niles!

Thanks. But how is it seen that C=2az_max?

… [itex]z_{max}[/itex] I believe is the longitudinal coordinate, where the velocity is zero.

v= 0 when z = z_max, so C = az_max

(the "2" would be for a different C)

 

[Niles]

Thanks. It is very kind of all of you to help me.Niles.

 

[Niles]

Is it correct that there are other ways to define C? E.g. as the initial velocity? In that case I would say that when z=0, then C=-0.5v²_initial, so I get
[tex]
2az = v^2 - v^2_{initial}
[/tex]

 

[tiny-tim]

Yes, C can be anything, and can be re-named later.

We often get C in an integration, then we tidy it up, and we find we have something awkward like 2π/C …

so we call that C (or v_o or z_o) instead! ​

(in this case, I think it was C/2)