Force as a function of velocity

  • Thread starter Chris2505
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  • #1
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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: [itex]F = ma[/itex] so [itex]\frac{F(v)}{m} = v \frac{dv}{ds}[/itex]

so: [itex]\int ds = m \int v \frac{dv}{F(v)}[/itex]

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

  • #2
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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....
 
  • #3
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s is the displacement, sorry. and using [itex]a = v \frac{dv}{ds}[/itex] you get the equation above. The Force only acts in one direcion and the force acts parallel to the velocity.
 
  • #4
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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?
 
  • #5
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If the force is constant, and so is mass, then why not use standard equations of motion?
 
  • #6
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I have an equation which describes force as a function of the velocity of the particle, and using [itex]\int ds = m \int v \frac{dv}{F(v)}[/itex] 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.
 
  • #7
306
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Well I suppose you could start with [tex]F(v)=m\frac{dv}{dt}[/tex] and do something similar.
 
  • #8
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ah, of course! Thanks! :)
 

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