# Differential equations particle motion

1. Sep 18, 2009

### philnow

1. The problem statement, all variables and given/known data
Hey all. I have F(v) and I'm looking for x(t).

F(v) = -bV2

I'm a physics undergrad student who hasn't done diff. equations in a while, so I'm very unsure of my work so far... but here it is.

3. The attempt at a solution

I write F=ma as F(v)=mdv/dt:

-bV2=mdv/dt

-m∫ dv/v2 = -b ∫dt

-m/v = -bt
m/v = bt

v = m/bt = dx/dt

∫ dx = m/b ∫ dt/t

Finally x(t) = m/b*ln(t)

Does this look good?

2. Sep 18, 2009

### rock.freak667

don't forget the constant.

Otherwise it looks alright. Also remember that ma can be written as mv dv/dx as well

3. Sep 18, 2009

### philnow

I forgot to add that the initial position and initial velocity are both zero so I don't need to worry about constants. Thanks!

4. Sep 18, 2009

### Redbelly98

Staff Emeritus
Are you sure the initial velocity is zero?

Differentiate that expression. Do you get dx/dt=0 at t=0???

5. Sep 19, 2009

### philnow

I misinformed you all, the particle has an initial velocity Vo. My bad. Even still I see that if I put 0 for t in my equation, it is undefined, when it should be zero.

I don't really know how to take it from here, any hints? How do I take into account the constant?

6. Sep 19, 2009

### Redbelly98

Staff Emeritus
This is where things go awry, since the equation is contradictory to v=v0 when t=0.

Yes, exactly. When you integrate, add a constant. What constant will give v=v0 when t=0?

7. Sep 19, 2009

### philnow

How does this seem?

x(t) = m/b*ln(t) + Vo*t

8. Sep 19, 2009

### Redbelly98

Staff Emeritus
Unfortunately, that does not give x(0)=0, due to the ln(t) term.

Also, the initial velocity appears to be infinite, again due to the ln(t) term.

If you show your work, we can help figure out what went wrong.

9. Sep 19, 2009

### philnow

Here's my work:

-bV2=mdv/dt

-m∫ dv/v2 = -b ∫dt

(note: so the integral of 1/v^2 is -1/v, and this integral is from initial v to final v) so:

bt = m(1/v - 1/vo) where vo is initial velocity

bt/m = 1/v - 1/vo

m/bt + vo = v

m/bt + vo = dx/dt

∫(m/bt + vo)dt = ∫dx

m/b*ln(t) + vo*t = x(t)

10. Sep 19, 2009

### Redbelly98

Staff Emeritus
Okay so far.

This step does not follow from the previous one. (We can tell because t=0 implies v=∞, instead of v=vo, in this equation.)

Instead, this step could have been

(bt/m) + (1/vo) = 1/v​
See if you can take if from there.

11. Sep 19, 2009

### philnow

(bt/m) + (1/vo) = 1/v

doesn't this mean v = m/bt + Vo?

12. Sep 19, 2009

### Redbelly98

Staff Emeritus
No, what it means is that v is the reciprocal of

(bt/m) + (1/vo)​

which is

$$\frac{1}{(bt/m) + (1/v_o)}$$

13. Sep 20, 2009

### philnow

Please tell me I've done this right:

from there:

∫dx = ∫https://www.physicsforums.com/latex_images/23/2356837-0.png [Broken]

x = m/b*ln(bt/m + 1/vo) - m/b*ln(1/vo)

for t=0, x=0 and when differentiated, at t=0,v=vo

Last edited by a moderator: May 4, 2017