Derivation of centripetal acceleration

To obtain an expression of a small body (i.e. a particle) describing circular motion, consider a body moving with constant speed v in circle of radius r. If it travels from A to B in a short interval of time δt then, since distance = speed*time, arc 'AB = vδt. Also, by the definition of an angle in radians arc AB = rδθ, therefore rδθ = vδt, which is the same as δθ = vδt/r.'

The vectors Va and Vb drawn tangentially at A and B represent the velocities at these points. The change in velocity between A and B is obtained by subtracting Va from Vb. That is: Vb - Va.

The line from the body to the centre of the circle turns through angle δθ, so this means the velocity direction turns through the same angle. The triangle formed between points A, B and the centre of the circle and the velcoity vector trangle have the same shape and the same angle between the two sides.

Provided this angle is small, then the distance between the two velocity vector lines, 'X' is equal to vδθ;

X = vδθ, however from the first equation δθ = vδt/r

X = v^2/r * δt.

Accleration is change in velocity over time = X/δt

Therefore a = v^2/r