# Equation of Motion

1. Dec 7, 2008

### kala

1. The problem statement, all variables and given/known data
A mass m has speed v0 at the origin and coasts along the x axis in a medium with force F(v). Use the chain rule of differentiation to write the equation of motion in the separated form m*v*dv/F(v)=dx.

2. Relevant equations
F(v)= -c(v^3/2)

3. The attempt at a solution
So far, i drew a diagram to show what was happening. I know that F(v) is the drag force. I'm really just confused about what the equation of motion is. I know that it is going to have a velocity and mass, I think.
Can any one help?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 7, 2008

### CompuChip

The equation of motion is just Newtons law,
$$F(v) = m a$$
where a is the acceleration
$$a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$.

3. Dec 7, 2008

### kala

Okay, that is what i have done so far, is this right,
I have m*v0= -c*v3/2.
then i am suppose to differentiate this using the chain rule which my book is calling the v*dv/dx rule to get a separated form m*v*dv/F(x)=dx. But i don't know what to differentiate with respect to, v or x, but my equation doesn't have an x. So maybe my equation is wrong, I am not quite sure.

4. Dec 7, 2008

### CompuChip

How do you get the right hand side? Don't use any specific form for F(v), just the equation I gave you.
Then, in m*a, replace a = dv/dt by something which looks like what you want (try dv/dt = dv/dt * something), using the chain rule.

5. Dec 7, 2008

### kala

So even though the drag force = F(v) i don't sub that part in?

6. Dec 7, 2008

### CompuChip

No, because there is also an F(v) and not c and v^(3/2) in your final answer.

7. Dec 7, 2008

### kala

so in the end when i use the chain rule should I end up with the drag force equation? Sorry I am still a little confused.

8. Dec 7, 2008

### CompuChip

No, they want you to show how you can go from
F(v) = m a
to
F(v) = m v (dv/dx)
using the chain rule, and then rewrite this to
m*v*dv/F(v)=dx.

The point being that the last line is a differential equation for v as a function of x with separated variables (all the v's on one side, all the x's on the other) so you can presumably solve it more easily than solving
m x''(t) = F(x'(t))
for x as a function of t.

9. Dec 7, 2008

### kala

ok, so understand what i was suppose to do, and got it to work, so was there any reason why they told me what the drag force was?

10. Dec 7, 2008

### CompuChip

Probably in the next question they are going to ask you to solve the equation by integrating both sides

11. Dec 7, 2008

### kala

they do ask me that, so do i substitute that in now for F(v)

12. Dec 7, 2008

### kala

never mind i totally got it to work! thank you for all of the help!