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 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 t²

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 t²

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_i² / a + (¹/₂) ( v² - 2 v v_i + v_i² ) / a

x - x_i = v v_i / a - v_i² / a + (¹/₂) ( v² - 2 v v_i + v_i² ) / a

x - x_i = - (¹/₂) v_i² / a + (¹/₂) v² / a

2 a (x - x_i ) = v² - v_i²

v² = v_i² + 2 a (x - x_i )

This provides the third of our "Big Three Kinetmatics Equations":

x = x_i + v_i t + (¹/₂) a t²

v² = v_i² + 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)?

Acceleration	Back to "Motion" Page		Free Fall

(c) Doug Davis, 2001; all rights reserved