# Integration Newtons Second Law

1. Feb 17, 2012

### Niles

1. The problem statement, all variables and given/known data
Hi

In a book they state
$$\frac{dv}{dt} = a \quad \Rightarrow \quad v(z) = \sqrt{2a(z-z_{max})}$$
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
$$\int vdv = \int adz \quad \Rightarrow\quad \frac{1}{2}v^2 + C = az$$
Here C denotes a constant. What should I do from here?

Best,
Niles.

2. Feb 17, 2012

### genericusrnme

what is z?

if $\frac{dv}{dt} = a$ then $\int dv = \int a\ dt \ \rightarrow v = \int a\ dt$

3. Feb 17, 2012

### tiny-tim

Hi Niles!
erm

you're there!

4. Feb 17, 2012

### Niles

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

5. Feb 17, 2012

### Pengwuino

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

6. Feb 17, 2012

### Niles

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

7. Feb 17, 2012

### tiny-tim

Hi Niles!
v= 0 when z = zmax, so C = azmax

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

8. Feb 18, 2012

### Niles

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

Best,
Niles.

9. Feb 18, 2012

### 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=-0.5v2initial, so I get
$$2az = v^2 - v^2_{initial}$$

10. Feb 18, 2012

### 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 vo or zo) instead!

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