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
vavg = (x - xi) / t
x = xi + vavg t
vavg = (v + vi) / 2
vavg = [ ( vi + a t ) + vi ] / 2
vavg = vi + (1/2) a t
Therefore,
x = xi + vi t + (1/2) a t2
We could start with our velocity equation,
and solve for the time t
t = ( v - vi ) / a
and then use that in our distance equation
x = xi + vi t + (1/2) a t2
x = xi + vi [ ( v - vi ) / a ] + (1/2) a [ ( v - vi ) / a ]2
x = xi + v vi / a - vi2 / a + (1/2) ( v2 - 2 v vi + vi2 ) / a
x - xi = v vi / a - vi2 / a + (1/2) ( v2 - 2 v vi + vi2 ) / a
x - xi = - (1/2) vi2 / a + (1/2) v2 / a
2 a (x - xi ) = v2 - vi2
v2 = vi2 + 2 a (x - xi )
This provides the third of our "Big Three Kinetmatics Equations":
x = xi + vi t + (1/2) a t2
v2 = vi2 + 2 a (x - xi )
If a vehicle brakes to a stop from 70 mi/hr in 186 feet, what is its acceleration (assumed to be constant)?
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(c) Doug Davis, 2001; all rights reserved