The question is: What ODE does v=dx/dt solve? And what is the limit as t approaches infinity of v(t). It sounds almost too easy, but I am confused about what to do exactly. Intuitively, as the question is about falling objects I would expect the limit to go to infinity but I am not sure which ODE is the correct one to give this answer...
Thanks a lot in advance!!
On
Do you mean for the velocity $v$ to be constant?
If not, then given the position as a function of time $x(t)$, we can compute the velocity at each time $t$ $$ v(t) = \dot{x}(t). $$ Solve the ODE involves integrating the velocity to find the position at each time $t$ in the domain: $$ \int^{t} v(\tau) d\tau = x(t) $$ In terms of falling, your time may be finite; it will impact at some point.
On
In the context of "falling objects" one may let $x$ denote the distance above the ground in some unit of distance such as meters and $t$ to denote the time in some convenient unit, such as seconds.
In such a situation the object is under a constant gravitational force $g$ which, since the object is falling, causes $x$ to decrease as time $t$ increases. In this situation (neglecting any atmospheric drag) $x$ will accelerate $\dfrac{d^2x}{dt^2}$ at a constant rate given by
$$ \frac{d^2x}{dt^2}=-g\tag{1}$$
The velocity of the object at time $t$ is the anti-derivative of the acceleration
$$v=\frac{dx}{dt}=-gt+v(0)\tag{2}$$
where $v(0)$ is the velocity at time $t=0$. If the object was merely dropped from some initial height $x(0)$, then $v(0)=0$. If it was thrown (either upwards or downwards) then $v(0)\ne0$.
The anti-derivative of $v(t)$ gives the height $x(t)$ above the ground at time $t$
$$x(t)=-g\cdot\frac{t^2}{2}+v(0)t+x(0)\tag{3}$$
where $x(0)$ is the initial height of the object above the ground (that is, when $t=0$.
In summary, in the context of falling objects, $v=\frac{dx}{dt}$ denotes the velocity of the falling object at some time $t$.
It is assumed that $\textbf{x}(t)$ is the particles position function and from first year calculus,
$$\frac{d\textbf{x}}{dt} = \lim_{h \to 0} \frac{\textbf{x}(t+h) - \textbf{x}(t)}{h}$$
However, the above also has the geometric interpretation as a displacement vector on the trajectory graph of $\textbf{x}(t)$ i.e the velocity vector and so we also call this quantity $\textbf{v}(t)$. In the image below, replace $\textbf{r}$ with $\textbf{x}$. The trajectory graph is the curved segment.
$$\textbf{v}(t):=\frac{d\textbf{x}}{dt} = \lim_{h \to 0} \frac{\textbf{x}(t+h) - \textbf{x}(t)}{h}$$
Hence the ODE is asking you to find such an $\textbf{x}$ given $\textbf{v}$. Your equation is separable, in which we know how to find the solution.
$$\textbf{v}(t) = \frac{d\textbf{x}}{dt} \Rightarrow \int \textbf{v}(t) \ dt = \int d\textbf{x} = \textbf{x}+ \vec{a}$$
where $\vec{a}$ is a constant vector.