Derivation of Centripetal Acceleration This is one of those things that most students often do not see it ever derived. The formula just appears in a textbook.

Consider the diagram below:

The object wants to move in a straight line from A to C and that distance is given by vt.

The object is also falling, in that time, a distance of 1/2 at² (a =2d/t²)

The motion from C to D, always directed towards the center, is the one we need to describe. Lets call the distance from C to D, x. With reference to the diagram above, the three relevant distances are:

• OA = radius of the circle = R
• AC = vt
• OC = R +x

So, courtesy of Phythagoras, we know that

R² + (vt)² = (R+x)²

R² + (vt)² = R² +2Rx + x²

(vt)² = x(2R+x)

If the time interval is very short then x << R, therefore

(vt)² = 2xR or x = 1/2(v²/R)t²

(v²/R) = 2x/t² = a