Derivation of kinetic energy

1. Feb 23, 2012

bentley4

Hi everyone,

There are 2 things I do not understand in the derivation of kinetic energy from work:

(1) W = $\int$$\vec{F}$(t).d$\vec{r}$(t)=

(2) m.$\int$$\frac{d\vec{v}(t)}{dt}$.d$\vec{r}$(t)=

(3) m.$\int$d$\vec{v}$(t).$\frac{d\vec{r}(t)}{dt}$=

(4) $\frac{m}{2}$.(v(t1) - v(t0))$^{2}$

Question I: I don't understand why you can just change the division like that between (2) and (3). I know multiplication and division have the same order but they are calculated from left to right. So why can you do that from (2) to (3)?
http://en.wikipedia.org/wiki/Order_of_operations

Question II: I don't understand why the change in kinetic energy between t1 and t0 is sometimes written as $\frac{1}{2}$.m.v(t1)$^{2}$ - $\frac{1}{2}$.m.v(t0)$^{2}$

After (4), it should be $\frac{m}{2}$.v(t1)$^{2}$ - m.v(t1).v(t0) + $\frac{m}{2}$.v(t0)$^{2}$ , right?

Last edited: Feb 23, 2012
2. Feb 23, 2012

Staff: Mentor

No. Your step (4) is incorrect.

3. Feb 23, 2012

bentley4

I'm sorry, I don"t understand what I am doing wrong going from (3) to (4). Can you elaborate please?

4. Feb 23, 2012

Staff: Mentor

∫v dv = v2/2 → v12/2 - v02/2

5. Feb 24, 2012

Philip Wood

I respect your unwillingness to regard $\frac{d\vec{v}}{dt}$ as one thing divided by another, but you might be happier to do this with $\frac{Δ\vec{v}}{Δt}$, and then associate Δt with Δ$\vec{r}$ to make $\vec{v}$. Only then take it to the limit.

I don't suppose this is rigorously valid, but I suspect this is how a lot of physicists justify this sort of transposition.

6. Feb 24, 2012

bentley4

7. Feb 25, 2012

juanrga

Question I: Because a derivative is equivalent to a division of differentials. And then by usual algebraic rules you can move the denominator.

Question II: It is not sometimes, but always, because that is the result of the integration of (3). You own result (4) is incorrect.

Last edited: Feb 25, 2012