Force as a function of velocity

[Chris2505]

I'm trying to find the final velocity and the time taken for a particle to tavel a certain distance. I know the initial velocity and displacement and the force as a function of the velocity.

Using: F = ma so \frac{F(v)}{m} = v \frac{dv}{ds}

so: \int ds = m \int v \frac{dv}{F(v)}

Rearanging and using limts gives final velocity as a function of dispacement and initial velocity. However I am unsure about how to calculate the time taken for this displacement?

 

[homology]

your force looks like its a function of 's' more than v. What is 's'? Is this the only force acting? Is the direction of motion parallel to the direction of the force...etc...

 

[Chris2505]

s is the displacement, sorry. and using a = v \frac{dv}{ds} you get the equation above. The Force only acts in one direcion and the force acts parallel to the velocity.

 

[homology]

I'm still missing the context for this. You won't be able to do anything without a specific form for F(v) and your limits of integration won't have times, they'll have distances and velocities. Are you trying to derive some general expression?

 

[_PJ_]

If the force is constant, and so is mass, then why not use standard equations of motion?

 

[Chris2505]

I have an equation which describes force as a function of the velocity of the particle, and using \int ds = m \int v \frac{dv}{F(v)} can calculate the displacement of the particle, using limits for the velocity.

Or, if re-arranged, the final velocity can be calculated if the change in displacement and initial velocity is known.

However what I also need to calculate is the time taken for the change in velocity. Which this equation can't do, is the an equation to calculate this?

Hope this makes more sense.

 

[homology]

Well I suppose you could start with F(v)=m\frac{dv}{dt} and do something similar.

 

[Chris2505]

ah, of course! Thanks! :)