- #1

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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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In summary, the conversation discusses the difficulty in finding velocity when the force is not constant but a function of distance. Integrating the equation for force does not account for the mass of the object, and there may be a need to use a quotient rule for integration. It is suggested that the mass of the object is constant and that the relevant integration property can be used to calculate velocity. However, there is another issue as velocity is the integral of acceleration with respect to time, not position. The chain rule may be helpful in finding the correct equation for velocity.

- #1

- 4

- 0

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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- #2

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DBaima22 said:

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:

[tex] \int g(x)\,dx = \int \frac{f(x)}{a}\,dx = \frac{1}{a}\int f(x)\,dx [/tex]

If something is constant (i.e. it has no dependence on the integration variable), then it can be brought outside the integral.

- #3

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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?

v= -1/3x^3+2.5x/mass?

- #4

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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. Hint: the chain rule may be of use here.

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