Acceleration is a derivative,
As with the velocity, we will usually mean the instantaneous acceleration if we simply say "the acceleration". We will often restrict ourselves situations with a constant acceleration; in that case the average acceleration is the same as the instantaneous acceleration.
We already know that
or
v_{avg} = (x  x_{i}) / t
x = x_{i} + v_{avg} t
v_{avg} = (v + v_{i}) / 2
v_{avg} = [ ( v_{i} + a t ) + v_{i} ] / 2
v_{avg} = v_{i} + (^{1}/_{2}) a t
Therefore,
x = x_{i} + v_{i} t + (^{1}/_{2}) a t^{2}
We could start with our velocity equation,
and solve for the time t
t = ( v  v_{i} ) / a
and then use that in our distance equation
x = x_{i} + v_{i} t + (^{1}/_{2}) a t^{2}
x = x_{i} + v_{i} [ ( v  v_{i} ) / a ] + (^{1}/_{2}) a [ ( v  v_{i} ) / a ]^{2}
x = x_{i} + v v_{i} / a  v_{i}^{2} / a + (^{1}/_{2}) ( v^{2}  2 v v_{i} + v_{i}^{2} ) / a
x  x_{i} = v v_{i} / a  v_{i}^{2} / a + (^{1}/_{2}) ( v^{2}  2 v v_{i} + v_{i}^{2} ) / a
x  x_{i} =  (^{1}/_{2}) v_{i}^{2} / a + (^{1}/_{2}) v^{2} / a
2 a (x  x_{i} ) = v^{2}  v_{i}^{2}
v^{2} = v_{i}^{2} + 2 a (x  x_{i} )
This provides the third of our "Big Three Kinetmatics Equations":
x = x_{i} + v_{i} t + (^{1}/_{2}) a t^{2 }
v^{2} = v_{i}^{2} + 2 a (x  x_{i} )
If a vehicle brakes to a stop from 70 mi/hr in 186 feet, what is its acceleration (assumed to be constant)?

 
(c) Doug Davis, 2001; all rights reserved