is s = 1/2at^2

Its velocity v at any time t is given by the first derivative of the distance function:

v = at

Its acceleration a at any time t is the derivative of the velocity, (or the second derivative of the distance)

a = a The acceleration is a constant.

Near the surface of the earth, a = 9.8 m/s^2 The derivative of a constant is zero (constants, by definition, do not change) so there is no second derivative, or jerk.

However, that is only an approximation for objects near the earth's surface; at great distances from the earth, the acceleration of gravity is less, but it increases as you approach the earth. For example, at a distance of 8000 miles from the center of the earth, or two earth radii, the gravitational acceleration is 1/4 what it is at the surface, or 2.45 m/sec^2. So an object dropped from that distance would experience a jerk because its acceleration would increase as it approached earth.

What's really amazing is that the purely mathematical procedure of "taking a derivative of a function" can be used to so accurately predict the motion of moving bodies in a gravitational field, a physical reality that exists independently of the logical structures mathematicians invent and play with.

Mathematically, the derivative is a procedure applied to a mathematical function which gives another function that tells you the rate of change of that function. And you can continue taking derivatives until you finally get to a constant, which has no change, so its derivative is zero.

I don't know about you, but I find that fucking amazing.

For those of you who haven't had calculus, the derivative of a polynomial function of the form y = mx^n is y' = nmx^(n-1). Granted, that expression didn't just come out of thin air, it is the result of a complex and clever mathematical proof. But the fact that it can be applied to the physics of falling bodies in the real world is just wonderful.