# More on Acceleration

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 can take the definition of acceleration, turn it around, and write

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

Sometimes we do not need the time but want a connection between distance, velocity, and acceleration.

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)?