# Velocity as a function of position x

1. Dec 9, 2013

### 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?

View attachment 64630

2. Dec 9, 2013

### vela

Staff Emeritus

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?

3. Dec 9, 2013

### 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

4. Dec 9, 2013

### KiNGGeexD

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

5. Dec 9, 2013

### vela

Staff Emeritus
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.

6. Dec 9, 2013

### 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:):)

7. Dec 9, 2013

### KiNGGeexD

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

8. Dec 9, 2013

### 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.

Last edited: Dec 9, 2013
9. Dec 9, 2013

### KiNGGeexD

Ok I understand, what a foolish mistake!!!

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

10. Dec 9, 2013

### KiNGGeexD

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

11. Dec 9, 2013

### 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.

12. Dec 9, 2013

### 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 ?

13. Dec 9, 2013

### 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.

14. Dec 9, 2013

### KiNGGeexD

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

15. Dec 9, 2013

### KiNGGeexD

So have I to use information like

dP/dt=F etc?

16. Dec 9, 2013

### KiNGGeexD

Would it be correct to say that

F(x)= m dv(x)/dt ??

17. Dec 9, 2013

### Office_Shredder

Staff Emeritus
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).

18. Dec 9, 2013

### KiNGGeexD

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

19. Dec 10, 2013

### 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

20. Dec 10, 2013

### vela

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