Just wanna make sure I have this right. The derivative of position is velocity, and the derivative of velocity is acceleration
so if $$F = ma$$
where $$a=\frac{d}{dx} v(t)\\ v(t)=\frac{d}{dx} s(t)$$ where $s(t)$ is the position in time.
This means that $$a = s''(t)$$ which means $$F = ms''(t)$$
and its an ordinary differential equation because it contains ordinary derivatives and it doesn't have the weird partial derivative symbol anywhere?
Is everything above right? Thanks.
On
Just expand a little on what everyone has already said: In general, assuming the mass $m$ and velocity $\mathbf{v}$ are functions of time only: $$ \mathbf{F}= \frac{d}{dt}\left(m\mathbf{v}\right) = m\frac{d \mathbf{v}}{dt}+ \frac{dm}{dt}\mathbf{v}$$ The above results follows from the fact that force is the rate of change of momentum with time, and momentum is the product of mass and velocity.
Now, if the mass $m$ is constant, then $\mathbf{v} =\frac{d \mathbf{s}}{dt}$ and $$\mathbf{F}= m\frac{d \mathbf{v}}{dt} = m\mathbf{a}$$ Finally (again assuming that $m$ is constant), if $\mathbf{s}$ is the displacement vector as a function of time, then $\mathbf{v} =\frac{d \mathbf{s}}{dt}$, which implies that $$\mathbf{F}= m\frac{d}{dt}\left(\frac{d \mathbf{s}}{dt}\right) = m\left[\frac{d^2}{{dt}^2}\mathbf{s}(t)\right] = m\mathbf{a}$$
Yes it is correct, in the particular case of a mass-spring system we have $F=-ks$ and the differential equation becomes
$$m\ddot{s}+ks=0$$