- #1

- 458

- 1

[tex]KE = \int_{0}^{s} \frac{d(mv)}{dt}ds = \int_{0}^{mv} v d(mv)[/tex]

I should really know this, but I don't. How do I get from the second expression to the third?

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- Thread starter dimensionless
- Start date

- #1

- 458

- 1

[tex]KE = \int_{0}^{s} \frac{d(mv)}{dt}ds = \int_{0}^{mv} v d(mv)[/tex]

I should really know this, but I don't. How do I get from the second expression to the third?

- #2

- 335

- 4

dimensionless said:

[tex]KE = \int_{0}^{s} \frac{d(mv)}{dt}ds = \int_{0}^{mv} v d(mv)[/tex]

I should really know this, but I don't. How do I get from the second expression to the third?

[tex]v \equiv \frac{ds}{dt}[/tex] is the definition of v. Since the integral is now an integral of d(mv) then the limits will go in terms of the new integration variable. (I am assuming we are speaking of an object that starts from rest and accelerates somehow to a speed v.)

-Dan

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