• Support PF! Buy your school textbooks, materials and every day products Here!

F(x,v) = kvx

  • Thread starter Berko
  • Start date
  • #1
68
0
I was given the following problem for a homework assignment:

Given F(x,v) = kvx where k is positive and the velocity is v-nought at x = 0, t = 0, what is x as a function of time?

First I derived v as a function of position:

v = v-nought + kx^2/2m

Then, I derived x as a function of time:

x = Sqrt[2mv-nought/k]tan(t*Sqrt[kv-nought/2m])

My problem with this solution is that x is then undefined for various points in time, plus the fact that it jumps from inifnity to negative infinity when it is undefined.

Now, if this is correct, it can be used as proof that the function as given cannot conform to anything in reality. However, it just may be plain out wrong. If it is wrong, I was wondering if someone might supply me with the correct answer as well as how to go about getting it.

Thank you very much for your time.
 
Last edited:

Answers and Replies

  • #2
James R
Science Advisor
Homework Helper
Gold Member
600
15
F(x,v) doesn't seem to tell us anything useful. It's just a random function, as far as I can see. Therefore, v is unconstrained, and x could be anything as a function of time.

Edit to add:

Ah...hang on. Is F a force?
 
  • #3
James R
Science Advisor
Homework Helper
Gold Member
600
15
If F = kvx is the force on the particle, then:

ma = kvx,

which means

[tex]m \frac{d^2 x}{dt^2} = kx \frac{dx}{dt}[/tex]

Can you solve that differential equation?
 
  • #4
68
0
Yes, it is a force, and I wrote kvx = ma = mv(dv/dx).

So, kx/m dx = dv, and

kx^2/2m = v - v-nought.

Therefore, v = v-nought + kx^2/2m = dx/dt.

I then seperated variables again and solved for x as a function of time, which resulted in my result written up in my first post, aloing with the problems it carries.
 
  • #5
James R
Science Advisor
Homework Helper
Gold Member
600
15
Ah, yes. I forgot that a = v(dv/dx).

I haven't checked your final answer, but it's probably correct. Your conclusion that the force is unrealistic seems sensible to me.
 
  • #6
68
0
Thank you.
 

Related Threads for: F(x,v) = kvx

Replies
1
Views
6K
Replies
10
Views
2K
  • Last Post
Replies
15
Views
886
Replies
7
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
13K
  • Last Post
Replies
17
Views
8K
Top