- #1

Steve Drake

- 53

- 1

I am confused by the momentum version of Newtons second law...

So we know

[tex]\bar{F}=m\bar{a}=m\left(\frac{d\hat{v}}{dt}\right) [/tex]

and that

[tex]\bar{\rho}=m\bar{v}=m\left(\frac{d\bar{x}}{dt}\right)[/tex]

so is

[tex]\frac{d\bar{p}}{dt}=m\frac{d\left(\frac{d\bar{x}}{dt}\right)}{dt}[/tex]

What I mean is this bit [tex]\frac{d\left(\frac{d\bar{x}}{dt}\right)}{dt}[/tex] somehow equal to [tex]\bar{a}[/tex]

Thanks