Velocity as a function of position x

[KiNGGeexD]

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

 

[vela]

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?

 

[KiNGGeexD]

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

 

[KiNGGeexD]

This is just a test to see if I can get my image to upload :)

 

[vela]

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.

 

[KiNGGeexD]

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:):)

 

[KiNGGeexD]

Sorry to be a pest but can I just confirm that my method up until the integration was correct?

 

[nasu]

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.

 

[KiNGGeexD]

Ok I understand, what a foolish mistake!

Typical student mentality i.e no thought lol! Thank you very much :)

 

[KiNGGeexD]

This is as far as I have gone? I'm not 100% what to do now?

 

[nasu]

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.

 

[KiNGGeexD]

I had an awful teacher for maths:( where did I go wrong??

I'm sure

a=F/m so a(x)=Fo+cx/m ?

 

[nasu]

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.

 

[KiNGGeexD]

So is there another method more suitable in which to solve it

 

[KiNGGeexD]

So have I to use information like

dP/dt=F etc?

 

[KiNGGeexD]

Would it be correct to say thatF(x)= m dv(x)/dt ??

 

[Office_Shredder]

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).

 

[KiNGGeexD]

Ahh ok thank you very much:)! That's really helpful!

 

[KiNGGeexD]

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

 

[vela]

It doesn't look right. What's M supposed to represent?

 

[KiNGGeexD]

Mass I didn't mean to put it in caps

 

[KiNGGeexD]

Ah it was to distinguish it between dm beausse I used d as a constant:)

 

[vela]

It still doesn't look right. The units don't work out.

 

[KiNGGeexD]

That's true!

What was the chain rule method you mentioned?

 

[vela]

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.

 

[nasu]

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.

 

[KiNGGeexD]

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

 

[nasu]

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.

 

[vela]

To me, it looks like you just made an algebra mistake, but it would help if you'd show your work.

 

[KiNGGeexD]

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

??