Velocity as a function of position x

AI Thread Summary
The discussion revolves around finding the velocity of a particle as a function of position x, given a force equation F(x) = Fo + cx. Participants emphasize the importance of integrating with respect to x rather than t, highlighting the need to express acceleration in terms of x. The correct approach involves using the chain rule to derive the relationship between velocity and position, leading to the integration of both sides of the equation. The conversation also touches on common mistakes in integration and the significance of handling sums correctly during the process. Ultimately, the participants guide the original poster towards the correct formulation of the velocity function.
KiNGGeexD
Messages
317
Reaction score
1
Hi all, I have completed a practice question and there are no model answers so I don't know if I'm correct in my method and answer! I have attached a photo, but here is what I'm doing.

Basically I have been asked to find the velocity as a function of x assuming that the particle has mass m at initial position x=0 and has a force (eqn below) acted on it

F(x)=Fo+cx

And a=F/m so use the force equation above and then integrate to get the velocity as a function of position?
https://www.physicsforums.com/attachments/64630
 
Physics news on Phys.org
KiNGGeexD said:
Hi all, I have completed a practice question and there are no model answers so I don't know if I'm correct in my method and answer! I have attached a photo, but here is what I'm doing.

Basically I have been asked to find the velocity as a function of x assuming that the particle has mass m at initial position x=0 and has a force (eqn below) acted on it

F(x)=Fo+cx

And a=F/m so use the force equation above and then integrate to get the velocity as a function of position?



https://www.physicsforums.com/attachments/64630
Your attachment didn't work.

Yes, you're supposed to integrate
$$a = \frac{dv}{dt} = \frac{1}{m}(F_0 + cx),$$ but did you integrate with respect to t or with respect to x?
 
I integrated with respect to dt?

My final solution was...v(x)= Fo+cx*t/m

That was the product of integrating with respect to dt, I assumed x to be a constant of sorts
 
This is just a test to see if I can get my image to upload :)
ImageUploadedByPhysics Forums1386610254.200680.jpg
 
You can't assume x is a constant; it varies with time. And integrating dv/dt with respect to t will give you v(t), not v(x). You need to express the acceleration a in terms of a derivative with respect to x. Hint: Use the chain rule.
 
Ahh that makes sense! To get v(x) I have to integrate with respect to dx?

Ok I shall give that a bash and get back with my result:)
Thanks a lot for your help:):)
 
Sorry to be a pest but can I just confirm that my method up until the integration was correct?
 
There isn't much before integration.
Yes, acceleration is F/m. That part is right.

You can also use work-energy theorem if you are not familiar with chain rule.
 
Last edited:
Ok I understand, what a foolish mistake!

Typical student mentality i.e no thought lol! Thank you very much :)
 
  • #10
ImageUploadedByPhysics Forums1386626081.213951.jpg


This is as far as I have gone? I'm not 100% what to do now?
 
  • #11
It's not easy to see but does not look good.
What is your math level? Did you learn about derivatives and integrals?
It seems that you are trying random things.

Do you know the work-energy theorem? I mean the relationship between the work done by a force and the change in kinetic energy.
 
  • #12
I had an awful teacher for maths:( where did I go wrong??

I'm sure

a=F/m so a(x)=Fo+cx/m ?
 
  • #13
Yes, this is OK, formally. But I don't see how you can use it to find v(x).
Unless you use the chain rule.
 
  • #14
So is there another method more suitable in which to solve it?:)
 
  • #15
So have I to use information like

dP/dt=F etc?
 
  • #16
Would it be correct to say thatF(x)= m dv(x)/dt ??
 
  • #17
King,

That last line is correct, but you need to be really careful when you are integrating.

v(t) = \int a(t) dt
is true, but
v(x) = \int a(x) dx
is not. For example if you accelerate at a flat 1 meter per second squared, then after one second you are traveling 1 m/s. After one meter you have actually been traveling for \sqrt{2} seconds, and your velocity after one meter is NOT equal to
\int a(x) dx = \int_{0}^{1} 1 dx = 1.If x is a function of t, x = x(t), then
F(x) = m \frac{dv}{dt} = m \frac{dv}{dx} \frac{dx}{dt} = m \frac{dv}{dx} v
So
v \frac{dv}{dx} = \frac{1}{m} (F_0 + cx )
and you have to solve this differential equation (which you can do by separation of variables).
 
  • Like
Likes 1 person
  • #18
Ahh ok thank you very much:)! That's really helpful!
 
  • #19
i got a rather complex answer which in the scope of what we have done in class

v(x)=√x(cx+2F)+dM/√m


where d is another constant
 
  • #20
It doesn't look right. What's M supposed to represent?
 
  • #21
Mass I didn't mean to put it in caps
 
  • #22
Ah it was to distinguish it between dm beausse I used d as a constant:)
 
  • #23
It still doesn't look right. The units don't work out.
 
  • #24
That's true!

What was the chain rule method you mentioned?
 
  • #25
It's what Office Shredder explained, that
$$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}.$$ If you check your calculus textbook, you'll see that's the chain rule.
 
  • #26
How did you integrate the right hand side of the equation? You should have an x^2 in it.
Do you know how to calculate a simple integral? The "d" in dx is neither constant nor variable. It's just a notation.
 
  • #27
Yea I understand calculus and integration and I know the chain rule, product rule, quotient rule etc and this may sound daft but I think it's the "as a function of" bit that is confusing me a little
 
  • #28
Then what is the problem?
In post 17 Office_Shredder pretty much solved it for you.
Just take the last formula in that post and integrate it to get v as a function of x.

You have dv and dx there.
Multiply both sides by dx and this will separate the variables. You will have vdv on the left and some (function of x)dx on the right.
So you integrate (over v) the left side and (over x) the right side, separately.
 
  • #29
To me, it looks like you just made an algebra mistake, but it would help if you'd show your work.
 
  • #30
Ok I have got myself confused I was thinking too much of the problem and did not realize how simple it is:)

All I do from

v*dv/dx= 1/m(Fo+cx)

Is multiply by dx so I have v(x)=(Fo+cx^2)/2m

??
 
  • #31
You didn't integrate correctly.
 
  • #32
KiNGGeexD said:
Ok I have got myself confused I was thinking too much of the problem and did not realize how simple it is:)

All I do from

v*dv/dx= 1/m(Fo+cx)

Is multiply by dx so I have v(x)=(Fo+cx^2)/2m

??
What do you get after just multiplying by dx? Before integration?
 
  • #33
vdv=1/m(Fo+cx) dx ?
 
  • #34
Or would it be

v dx dv=1/m(Fo+cx) dx
 
  • #35
KiNGGeexD said:
vdv=1/m(Fo+cx) dx ?

OK. So far so good.

Then what is \int vdv?
And what is \int \frac{1}{m} (F_o+cx) dx?
 
  • #36
vdv=(Fo+cx)/m dx

So

v^2/2= (Fo+x^2)/2m

So

v(x)= sqrt of the above lol
 
  • #37
Not right yet.
The integral over v is OK.
The other one is not.
You have a sum there but you integrate only one term of the sum. Kind of. The 1/2 goes only to the last term.

So, how do you integrate a sum of terms?
 
  • #38
(Fox+x^2)/2mx
 
  • #39
Are you trying random guesses again?
How do you deal with a sum, when you need to integrate? You did not answer this.
Suppose you have to integrate f(x)=a+bx. What is the integral of f(x)?
 
Last edited:
  • #40
I may word this odd but it is that the integral of the sum is the sum of the separate integrals of the separate components?

So (Fo+cx) dx

Is the same as

Fo dx + cx dx

?
 
  • #41
Yes. This is correct.
Now integrate each piece separately.
 
  • #42
Ok thank you:)!
 
  • #43
So

v=sqrt of Fo*x+cx^2/m

??
 
  • #44
If you don't use any parentheses, what you wrote reads like this:
v=\sqrt{F_o \cdot x +\frac{cx^2}{m}}
Is this what you mean?
If yes, it's not right. But it's getting closer. :)
I wonder why won't you write it step by step rather than throwing shots in the dark?
 
Last edited:
  • #45
It wasn't a shot in the dark I integrated

Fo dx

So I get Fo*x?

Integrate cx and I get cx^2/2

I'm really not sure where I am going wrong?
 
  • #46
Fo*x+cx^2/2m

Is what I get before isolating v

Which is v^2/2...
 
  • #47
Isn't the Fo term divided by m as well?
 
  • #48
Yea I meant for the whole thing to be divided by m... Sorry it's my phone and the way I have to type isn't ideal sorry if I'm causing any frustration:(
 
  • #49
Oh, I was afraid that you may get frustrated.:smile:
OK, the whole thing is divided by m. But not by 2. No matter how you interpret it, it is not completely right.
So it should be
\frac{v^2}{2}=\frac{F_o \cdot x +\frac{cx^2}{2}}{m}
or
\frac{v^2}{2}=\frac{F_o \cdot x}{m} +\frac{cx^2}{2m}
Right?
Have you got this, so far?
 
  • #50
Yea haha!

What I wrote before was before I isolated for the v term:)

I will need to get back to you on those formula because it is hard to decipher on my phone:(
 

Similar threads

Finding acceleration from Velocity vs Position graph
Replies
2
Views
4K
Vector is a function of its position or not?
Replies
16
Views
1K
Kinematics: find the particle's position as a function of time
Replies
4
Views
2K
Will The Cars Stop In Time? - Check My Work Please :)
Replies
13
Views
1K
Initial velocity of bird landing on horizontal wire modeled as spring
Replies
7
Views
974
Slight Confusion in Introductory Mechanics (Dynamics Problem)
Replies
15
Views
733
Relation between Friction, Velocity, and Acceleration
Replies
16
Views
984
The gravitating of a small mass towards a big mass
Replies
3
Views
2K
Irodov- 1.123- Simple conceptual question with two masses connected by a spring
Replies
20
Views
638
Find force as a function of position: F=F(x) using v=v(t)
Replies
6
Views
990

Hot Threads

Recent Insights

Back
Top