# Equation of motion with proportional drag

1. Dec 19, 2012

### Hannibal123

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

the forces will be like this

$$m⋅dv/dt=-m⋅g-k⋅v$$

I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0

3. The attempt at a solution
i have found this
$$v(t)=mg/k*(-1+e^(-k/m⋅t))+v_0⋅e^(-k/m⋅t)$$ or in picture form http://imgur.com/WjksG

But I am not sure it's correct

2. Dec 19, 2012

### ehild

It looks correct, but v0=0 .

Edit: I misread vo, so it is correct.

ehild

Last edited: Dec 19, 2012
3. Dec 19, 2012

### haruspex

4. Dec 19, 2012

### haruspex

I read the OP condition as initially (t,v) = (0, v0)

5. Dec 19, 2012

### ehild

Sorry, I misread it. Edit my post.

ehild

6. Dec 19, 2012

### Hannibal123

that is what i meant. I doubt the results because when i draw them in a graph they dosen't seem to be correct. I have drawn them i geogebra if you are familiar with that software.
My values are:
m: 0,145 kg
k: 0,0032
g: (gravity acceleration) 9,82
v_0: 9,93 m/s

this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
if you want to try for yourself (should save you some time). These values are not meant for a proportional drag, but still it seems weird that the graph looks like one of a constant function
http://imgur.com/FFElN (the red one)

7. Dec 19, 2012

### Hannibal123

Furthermore when i integrate the velocity equation i get this http://imgur.com/IO6G6
Again not sure if I am correct

8. Dec 19, 2012

### haruspex

How did -k/m end up as 0 / 0.15? Looks like rounding error, and the red line graph seems to be a consequence.

9. Dec 19, 2012

### Hannibal123

10. Dec 19, 2012

### Hannibal123

this is the equation of place (if that is the correct term in english) as far as I am concerned the height is not suposed to be negative to positive time values???
http://imgur.com/CxAYC

11. Dec 19, 2012

### haruspex

Your velocity time graph shows velocity +ve at time 0, so the distance time graph should show distance increasing at time 0.

12. Dec 20, 2012

### Hannibal123

Indeed it should, however it dosen't. Is this a wrong integration of the velocity equation?
http://imgur.com/A6BKT
Based on the gaph it does not seem to be correct.

13. Dec 20, 2012

### Hannibal123

Just found the correct place equation i think
http://imgur.com/quiLu
can you see a good way to shoten it?

14. Dec 20, 2012

### haruspex

Looks right. I'd write it as
$\frac{mg}{k}\left(-t+\left(\frac{m}{k}+\frac{v_0}{g}\right)\left(1-e^{-\frac{kt}{m}}\right)\right)$