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: $s\left(t\right)={t}^{3}-3{t}^{2}-5t+6$

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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