Consider a particle moving in a circular path in anti clockwise direction. Let the initial position of the particle be A and the final position be B. Let the initial and final position vectors be vector r and vector r¹ respectively. Similarly, the velocities at points A and B are vector v and vector v¹ which are tangents to the point A and B respectively.  Though the magnitudes of vectors v and v¹ are the same, the directions are different.  Let ∆θ be the angle subtended by AOB.

Compare the radius and the velocity vectors. By magnitude, r and r¹ vector are same. Similarly, v and v¹ have the same magnitude. The vector r and v are perpendicular to each other. Similarly, r¹ and v¹ are perpendicular to each other. The angle between r and r¹ is same as angle between v and v¹. Hence, we can draw a triangle PQR to represent our understanding.

By the property of similar triangle,  ∆v/v = ∆r/r

 ∆v= ∆r v/r

Dividing throughout by ∆t, ∆v/∆t = ∆r/∆t v/r

Acceleration, ∆v/∆t = v/r velocity (∆r/∆t)

Accelaration = v² / r which gives the magnitude of centripetal acceleration

Consider Acceleration= ∆v/∆t ; As ∆t does not have any direction, direction of centripetal acceleration depends on ∆v. This is directed towards the centre of the circle O as shown in the diagram.