A previous example of velocity used the DIFFERENCE QUOTIENT method to solve for velocity of a position function. Here we use the derivative (which is the the DIFFERENCE QUOTIENT technique when the increment approaches 0) to find the position function (to find velocity) AND the second derivative of the position function to find acceleration (the rate of change of velocity).
If 's' is distance in feet, find the instantaneous velocity and acceleration at t = 3 seconds for the function: $s\left(t\right)={t}^{3}-3{t}^{2}-5t+6$
Velocity and Acceleration

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