Position velocity acceleration
Position velocity acceleration
We now know that taking the derivative of a function will give us the slope, or the instantaneous rate of change of the function. So what if we take the derivative of a function that models the position of some object moving along a line? It gives us its velocity! And if we differentiate its velocity function? It gives us its acceleration! In this section, we will study the relationship between position, velocity and acceleration using our knowledge of differential calculus.
Lessons
Notes:
motion along a straight line
$s(t)$: position
$v(t)=s'(t)$: instantaneous velocity
$a(t)=v' (t)=s''(t)$: acceleration

1.
The position of a particle moving along a straight line is given by:
$s(t)=t^312t^2+36t14$
where t is measured in seconds and s in meters.