Calculus. The derivative of y with respect to x is the slope of a line tangent to the curve of y at point x, or the rate of change in y at point x of the function y(x). The rate of change of distance is simply velocity. Therefore the derivative of distance (as a function of time) is velocity. The rate of change in velocity is simply acceleration (once again as a function of time), so the derivative of velocity is acceleration. Now, the derivative of y with respect to x of y = x^n is n*x^(n-1). So if we start with x^n and want to find a function whose derivative is x^n. the derivative of x^(n+1) is (n+1)x^n, so the derivative of (x^(n+1))/(n+1) is x^n (since the derivative of some constant * f(x) is just that constant times the derivative of f(x). Knowing this, we can move backwards from the acceleration to velocity. If we have a constant acceleration A, then we know that the derivative of our velocity function = A. We can think of this as A*t^0, since t^0 = 1. So the function whose derivative is A is A*t^(0+1) divided by (0+1), which equals A*T. Now, the derivative of a constant is 0, so we have to add a constant to our velocity function to account for the range of possible functions. So velocity equals A*T + C1. We solve for this constant by setting the velocity at some point to some value, called V0. We usually set it equal to this value when time equals 0 because that's the point that our function "starts". So v0 = A*0 + C1, so v0 = c1 so velocity equals A*T + V0. Now we do the same thing to find displacement as a function of time. A*T^(1+1)/(1+1) + v0*T^(0+1)/(0+1) = 0.5A*T^2 + v0*T + C2 = displacement! solve for the constant again to get c2 = d0 = initial displacement.
