## Wednesday, February 1, 2017

### Displacement and Acceleration

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 = (V + Vi) / 2

where Vf is final velocity, Vi is initial velocity, and Vave is the average velocity.

df = di + vt

Understanding that v is average velocity, the equation becomes:

df = di + ((V + 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 + at

Step 2.) Find the average velocity: Vave = (V + Vi) / 2

Step 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.