Is this equation solvable for v?

  • B
  • Thread starter Intle
  • Start date
  • #1
27
0

Main Question or Discussion Point

Hello
I am attempting to solve this equation (a physics project); however, I seem to be getting stuck in a cycle between attempting to undo the square root and then dealing with the resulting quadratic. I have run out of creative solutions to get the variable v by itself. Any help would be greatly appreciated.

F = (m/(1-v2/c2)-1/2) * (v/t-b/t)
 

Answers and Replies

  • #2
201
51
By squaring both sides I can isolate a quadratic equation in v but it's huge.... Did you try that?
 
  • #3
mathman
Science Advisor
7,771
419
As written the equation looks funny. You have an expression raised to power -1/2 in the denominator. Why not +1/2 in the numerator?
 
  • #4
27
0
As written the equation looks funny. You have an expression raised to power -1/2 in the denominator. Why not +1/2 in the numerator?
Well you could rearrange the equation to be

F * (1-v^2/c^2)^1/2 = m * (v/t - b/t)
but i do not understand how that will really solve anything. If I now square both sides to get rid of square root over (1-v^2/c^2) I now end up with a v^2 -2vb + b^2 on the other side which is an issue because now how do I get the v byitself over here without messing up the other side again. That is the issue I have been happening. by the way, is there a better way to write these equations. I feel as though the way I have been typing this equation in text format may be a bit confusing?
 
  • #5
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,438
5,180
Well you could rearrange the equation to be

F * (1-v^2/c^2)^1/2 = m * (v/t - b/t)
but i do not understand how that will really solve anything. If I now square both sides to get rid of square root over (1-v^2/c^2) I now end up with a v^2 -2vb + b^2 on the other side which is an issue because now how do I get the v byitself over here without messing up the other side again. That is the issue I have been happening. by the way, is there a better way to write these equations. I feel as though the way I have been typing this equation in text format may be a bit confusing?
What's the problem? If you square both sides you get a quadratic in ##v##. The coefficients are a bit messy, but there's nothing you can do about that.
 
  • #6
201
51
Well you could rearrange the equation to be

F * (1-v^2/c^2)^1/2 = m * (v/t - b/t)
but i do not understand how that will really solve anything. If I now square both sides to get rid of square root over (1-v^2/c^2) I now end up with a v^2 -2vb + b^2 on the other side which is an issue because now how do I get the v byitself over here without messing up the other side again. That is the issue I have been happening. by the way, is there a better way to write these equations. I feel as though the way I have been typing this equation in text format may be a bit confusing?
You can use LaTeX to write the equations. Using LaTeX your equation is [tex] F = \frac{m (\frac{v}{t} - \frac{b}{t})}{\sqrt(1 - \frac{v^2}{c^2})} [/tex] A guide to LaTeX can be found here and here.

Regarding your equation. Squaring it you get: [tex] F^2 = \frac{m^2 (\frac{v^2}{t^2} + \frac{b^2}{t^2} - 2\frac{vb}{t^2})}{(1 - \frac{v^2}{c^2})} [/tex]

Algebraically manipulating should get you a quadratic in v.
 
Last edited:
  • #7
27
0
You can use LaTeX to write the equations. Using LaTeX your equation is [tex] F = \frac{m (\frac{v}{t} - \frac{b}{t})}{\sqrt(1 - \frac{v^2}{c^2})} [/tex] A guide to LaTeX can be found here and here.

Regarding your equation. Squaring it you get: [tex] F^2 = \frac{m^2 (\frac{v^2}{t^2} + \frac{b^2}{t^2} - 2\frac{vb}{t^2})}{(1 - \frac{v^2}{c^2})} [/tex]

Algebraically manipulating should get you a quadratic in v.
Okay, thanks for the LaTex guide. Now could you show me how you would algebraically solve it from where you stopped? I still end up getting stuck.
 
  • #8
mathman
Science Advisor
7,771
419
Okay, thanks for the LaTex guide. Now could you show me how you would algebraically solve it from where you stopped? I still end up getting stuck.
Write out the equation at the point you are stuck. To me it is a simple quadratic.
 
  • #9
201
51
Okay, thanks for the LaTex guide. Now could you show me how you would algebraically solve it from where you stopped? I still end up getting stuck.
That would be solving the equation for you. Like @mathman said why don't you show us where you're getting stuck?! And we'll help you out.
 
  • #10
27
0
That would be solving the equation for you. Like @mathman said why don't you show us where you're getting stuck?! And we'll help you out.
Okay, no problem. I'll bold the variable v.

First I multiply both sides by the denominator 1- v^2/c^2 and distribute the m^2 to get

F^2 -(F^2*v^2)/c^2 = (m^2*v^2)/t^2 + (m^2*b^2)/t^2 - (2*v*b*m^2)/t^2

Now I put all the terms with v on one side.

F^2 -(m^2*b^2)/t^2 = (m^2*v^2)/t^2 + (F^2*v^2)/c^2 - (2*v*b*m^2)/t^2

This is were I get stuck. I can't pull out any like terms , since not all the v are squared I can't pull them out either. This is the point where I'm stuck. I'm probably just overlooking something but I would appreciate it if you could now show me where my error is.
 
  • #11
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,438
5,180
Okay, no problem. I'll bold the variable v.

First I multiply both sides by the denominator 1- v^2/c^2 and distribute the m^2 to get

F^2 -(F^2*v^2)/c^2 = (m^2*v^2)/t^2 + (m^2*b^2)/t^2 - (2*v*b*m^2)/t^2

Now I put all the terms with v on one side.

F^2 -(m^2*b^2)/t^2 = (m^2*v^2)/t^2 + (F^2*v^2)/c^2 - (2*v*b*m^2)/t^2

This is were I get stuck. I can't pull out any like terms , since not all the v are squared I can't pull them out either. This is the point where I'm stuck. I'm probably just overlooking something but I would appreciate it if you could now show me where my error is.
I thought you were aiming at a quadratic? A quadratic in ##v## has terms in both ##v## and ##v^2## plus a constant term. That's what you've got, isn't it? All you have to do is put the two ##v^2## terms together.

What you have is of the form:

##c = a_1 v^2 + a_2 v^2 - bv##

And that's a quadratic.
 
  • #12
27
0
I thought you were aiming at a quadratic? A quadratic in ##v## has terms in both ##v## and ##v^2## plus a constant term. That's what you've got, isn't it? All you have to do is put the two ##v^2## terms together.

What you have is of the form:

##c = a_1 v^2 + a_2 v^2 - bv##

And that's a quadratic.
I was so bent on finding a direct equation I completely missed that solution. Thanks for all the help.
 

Related Threads on Is this equation solvable for v?

Replies
12
Views
1K
  • Last Post
Replies
17
Views
3K
Replies
7
Views
2K
Replies
7
Views
1K
Replies
7
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
12
Views
2K
Replies
78
Views
54K
Replies
1
Views
1K
Top