How do I find velocity when force is not a constant?

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SUMMARY

The discussion focuses on calculating velocity when force is a function of distance, specifically using the equation F(x) = 2.5 - x². The user initially attempted to integrate the force equation but failed to account for mass, leading to an incorrect velocity expression. A key insight provided is that while mass can be treated as constant, velocity must be derived by integrating acceleration with respect to time, not distance. The chain rule is suggested as a necessary tool for correctly determining velocity in this context.

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  • Understanding of Newton's second law (F=ma)
  • Basic knowledge of calculus, specifically integration
  • Familiarity with the chain rule in calculus
  • Concept of constant mass in physics
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DBaima22
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how can i find velocity if force is not constant, but a function of distance. I tried integrating the equation for the force, but that doesn't account for the mass of the object.

the given equation for the force is F(x)=2.5-x^2, where x is distance and since F=ma I have
a=(2.5-x^2)/m.

When I tried integrating it I got velocity= -1/3x^3+2.5x which does not account for mass

Is there a quotient rule for integration?
 
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Welcome to PF, DBaima22 :smile:
DBaima22 said:
how can i find velocity if force is not constant, but a function of distance. I tried integrating the equation for the force, but that doesn't account for the mass of the object.

the given equation for the force is F(x)=2.5-x^2, where x is distance and since F=ma I have
a=(2.5-x^2)/m.

When I tried integrating it I got velocity= -1/3x^3+2.5x which does not account for mass

Is there a quotient rule for integration?

It's completely reasonable to assume that the mass of the object is constant. Therefore, this becomes trivial. You can use the relevant integration property, namely that if g(x) = f(x)/a where a = const. then:

\int g(x)\,dx = \int \frac{f(x)}{a}\,dx = \frac{1}{a}\int f(x)\,dx

If something is constant (i.e. it has no dependence on the integration variable), then it can be brought outside the integral.
 
So, just to be clear, if acceleration is (2.5-x^2)/m. I can write velocity as
v= -1/3x^3+2.5x/mass?
 
DBaima22 said:
So, just to be clear, if acceleration is (2.5-x^2)/m. I can write velocity as
v= -1/3x^3+2.5x/mass?

Yes. BUT there is another problem with your solution that I only just realized (because I wasn't paying close enough attention before). You CAN'T get velocity by integrating acceleration with respect to position. Velocity is the integral of acceleration with respect to TIME. So you'll have to work a little bit harder to figure out how to get velocity. :wink: Hint: the chain rule may be of use here.
 

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