- #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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- Thread starter DBaima22
- Start date

- #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?

- #2

cepheid

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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?

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

cepheid

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Science Advisor

Gold Member

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- 38

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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