I have a bidimensional circunference centered in (5,5) made of 1000 points. So the equation that describes the circunference is: $$(x-5)^2 + (y-5)^2 = 25$$
Imagine that I have a vehicle that is running in that circunference as reaches 1 point for each second. How can I get the speed vector (x and y) and the acceleration (x and y)?
Hint:
I suppose a uniform circular motion. In this case the angular velocity is $$\omega=\frac{2\pi}{1000} $$ so the equations of motion are: $$ x=5+5\cos\left(\frac{2\pi}{1000}t \right) \qquad y=5+5\sin\left(\frac{2\pi}{1000}t \right) $$
deriving these you can find the equations for the velocity and the acceleration.
For uniform (constant speed) circular motion, the speed, $v$ is the circumference divided by the period. The circumference is 10$\pi$ (in the same length units as $x$ and $y$) and the period is 1000 (seconds). The direction of the velocity vector is perpendicular to the radius vector from the center of the circle to point $(x,y)$, so $\vec{v}$ is parallel to $(y-5)\hat{i}-(x-5)\hat{j}$. Therefore $$\vec{v} = \pm \frac{10\pi}{1000}\frac{(y-5)\hat{i}-(x-5)\hat{j}}{\sqrt{(x-5)^2+(y-5)^2}} = \frac{\pm \pi}{500}[(y-5)\hat{i}-(x-5)\hat{j}]$$ The $\pm$ sign indicates the velocity could be either clockwise (+) or counter-clockwise (-).
The acceleration has magnitude $v^2/r$ and its direction is opposite to the radius vector. Therefore $$\vec{a} = \frac{-\pi^2}{5(500)^2}\frac{(x-5)\hat{i}+(y-5)\hat{j}}{\sqrt{(x-5)^2+(y-5)^2}} = \frac{-\pi^2}{5^2(500)^2}[(x-5)\hat{i}+(y-5)\hat{j}]$$
