The acceleration in circular motion is [b]centripetal[/b], i.e. directed toward the center of the trajectory.

Move either point A or B such that they are close to each other. [math]v1[/math] and [math]v2[/math] are the velocities of the point, respectively, in A and B. [math]v2'[/math] is equal to [math]v2[/math], but is drawn such that it starts from A, so we can compute graphically the difference [math]dv=v2-v1[/math].[br][br][math]dv[/math] is always perpendicular to the base of an isosceles triangle formed by [math]v1[/math], [math]v2[/math] and [math]dv[/math]. The height of such a triangle is parallel to the displacement x(B)-x(A). The vector a is compute as [math]dv/alpha[/math]: when [math]alpha[/math] goes to zero, the time interval dt goes to zero, too, being [math]alpha=omega dt[/math].When A and B coincide, the acceleration, whose length is constant, points toward the center of the trajectory.