# Equation of motion integration

1. Dec 11, 2008

### Perillux

I'm just reading a basic physics tutorial, here is what it says in the section this post concerns:

I don't get how the go from the second equation to the third equation (the one with Vo). I do know how to integrate, but what are they taking the integral with respect to?

Last edited by a moderator: Mar 20, 2009
2. Dec 11, 2008

Think of it as multiplying the the dt over to the other side, and then dividing the v the other way. So we have.

-$$\lambda$$dt=m*dv/v

Now integrate both sides, one is an integral of dt the other dv.

Mathematically, this is not exactly what's going on, but you'll learn more when you take a differential equations course. Try to figure out how they get the v_0... it's from the constant of integration.

3. Dec 11, 2008

### Perillux

I'm taking a differential equations class next semester.

So, I get this: -$\lambda$t + C = m*ln(v) + C
then I can treat this as a single constant and I divide by m. Then I can raise (e) to the power of both sides and I have
$$e^{-\lambda t/m + C} = v$$
which can be written:
$$e^{C}e^{-\lambda t/m} = v$$

and since $e^{C}$ is a constant I guess that is where V_0 comes from ??

4. Dec 11, 2008

### Dr.D

The original statement was, "Newton's second law is insufficient to describe the motion of a particle." The discussion then degenerated into a particular example a particle subject to viscous friction only. The example essentially shifted the focus away from the original statement. The point of the original statement was just this:

Until you know how to express the forces involved, Newton's second law only provides the frame work for expressing the relation between motion and forces. The forces must be expressed separately if Newton's second law is going to be used to determine the motion.

5. Dec 11, 2008

### Perillux

Thanks for clarifying that. I knew that what I was asking wasn't vital to understanding what they are saying, but I still like to know how they got there.