Differential equations particle motion

[philnow]

Homework Statement

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.

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?

 

[rock.freak667]

Homework Statement

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.

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 +constant

don't forget the constant.

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

 

[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!

 

[Redbelly98]

Are you sure the initial velocity is zero?

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

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

 

[philnow]

I misinformed you all, the particle has an initial velocity V_o. 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?

 

[Redbelly98]

-m∫ dv/v² = -b ∫dt

-m/v = -bt

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

don't forget the constant.

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

 

[philnow]

How does this seem?

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

 

[Redbelly98]

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.

 

[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)

 

[Redbelly98]

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

Okay so far.

m/bt + vo = v

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.

 

[philnow]

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

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

 

[Redbelly98]

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

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

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

(bt/m) + (1/vo)​

which is

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

 

[philnow]

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

(bt/m) + (1/vo)​

which is

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

Please tell me I've done this right:

from there:

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

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