# Force as a function of velocity

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 im unsure about how to calculate the time taken for this displacement?

## Answers and Replies

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....

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.

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?

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

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.

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

ah, of course! Thanks! :)