If the relationship between displacement, acceleration and elapsed time is extended to the most general case, the following equations need to be combined.

1.)

df = di + vt

where

*df*is the final displacement from the fixed point,*di*is the initial displacement from the fixed point,*v*is rate of change in position, (average velocity), and*t*is elapsed time.
2.)

Vf = Vi + at

where

*Vf*is final velocity,*Vi*is initial velocity,*a*is the rate of acceleration, and*t*is the elapsed time.
3.)

Vave = (Vf + Vi) /2

where

*Vf*is final velocity,*Vi*is initial velocity, and Vave is the average velocity.
To begin, start with the general displacement equation:

*df = di + vt*

Understanding that v is average velocity, the equation becomes:

*df = di + (*(

*Vf + Vi) /*2) • t

And since

*Vf*can be found in relationship to acceleration, the following emerges:*df = di + (*((

*Vi + at)*

*+ Vi) /*2) • t

Solving the problem in steps generally is more easily understood. To find displacement when acceleration is present, do the following.

Step 1.) Find the final velocity:Vf = Vi + atStep 2.) Find the average velocity: Vave = (Vf + Vi) /2Step 3.) Find the displacement:df = di + vt

EXAMPLE

A model car is 3 meters from the starting line on a model car race track and it is moving at 2 m/s. It accelerates at a rate of 5 (m/s)/s for 4 seconds. How far does it end up from the starting line?

Step 0.) Collect the data!

di = 3 m

vi = 2 m/s

a = 5 (m/s)/s

t = 4

Step 1.) Find final velocity.

Vf = vi + at

vf = 2 + 5•4

vf = 2 + 20

vf = 22

Step 2.) Find average velocity.

v(ave) = (vf + vi)/2

v(ave) = (22 + 2)/2

v(ave) = 24/2

v(ave) = 12 m/s

Step 3.) Find the total displacement.

df = di + vt

(remember v is average velocity and t is the same elapsed time as for the acceleration)

df = 3 + 12•4

df = 3 + 48

df = 51 meters

So, the model car ends up 51 meters from the starting line.

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