In famous quantum physics book of Cohen-Tannoudji, the authors write in the conclusion of the first chapter, that in classical physics, one needs both the vector $\vec{r}(t)$ and $\vec{v}(t)$ to describe a particle.
But why don't we only need $\vec{r}(t)$ since the vector of speed could be obtained by the derivation of the position ?
In Newtonian Mechanics, we basically solve the equation: $$F(x,t)= m \frac{d^2x} {dt^2}$$ Since this is a second-order differential equation, you need two boundary conditions. In the Newtonian framework, boundary conditions are initial position and initial velocity.
So putting the values you get the trajectory of the particle.
Suppose, you get the trajectory of the particle as $y=f(x,t)$, Then you can get velocity as $v= \frac{dx}{dt}=f'(x,t)$
So knowledge of $v(t)$ is implicit in classical mechanics. without the knowledge of $v$ you don't have analytical knowledge of position, so you can't find out the velocity.
Also, In other formulations such as Lagrangian and Hamiltonian formulation position and velocity are very much intertwined. Also, there is well-built concept of phase space, that I am not discussing.
Finding the trajectory in real space is not at all easy as it seems. Thats why concept of lagrangian formulation and phase space have been developed.
But why don't we only need $\vec{r}(t)$ since the vector of speed could be obtained by the derivation of the position ?
The way we express it is not perfectly clear, but the authors are saying that we need only $\vec{r}(t)$ and $\vec{v}(t)$ at one instant in time in order to find $\vec{r}(t)$ for all $t$.
If you already had $\vec{r}(t)$ for all $t$, then of course you could find $\vec{v}(t)$. But also, you wouldn't have any problem left to solve.
Of course if $\mathbf x = \mathbf f(t)$ is known, we don't need more information.
The problem is how to determine $\mathbf f(t)$ knowing $\mathbf F(\mathbf x)$ (force as a function of position)? We need $\mathbf x(t_0)$ and $\mathbf v(t_0)$ for some $t_0$.
