Gravitational Acceleration: Google search yields only incorrect
analyses

Gravitaional acceleration 9.8 meters/sec^2
approx. 10 meters / sec^2.
rate of change in velocity(secs)= 20 m/s under gravitational
acceleration

time total distance final velocity average velocity
1sec 10m --- 10 m/s
2 40 40 20
3 90 60 30
4 160 80 40
5 250 100 50
6 360 120 60
7 490 140 70

At the end of 4 secs, for example the final velocity is 80 m/s. It has
traveled 160 m .

Gravitational Acceleration: Google search yields only incorrect
analyses

Gravitational Acceleration: Google search yields only incorrect
analyses

Gravitational Acceleration: Google search yields only incorrect
analyses

Acceleration is a derivative of a derivative. Velocity is, a rate of
distance in time.

Acceleration then is, rate of change in the rate of
distance in time. The acceleration of gravity is, the distance change
as a function of time. cm/sec^2.

If an object goes from 0 to 50 ft per second in 5 seconds, it's
acceleration is 50 ft per second / 5 seconds.

This simplifies to
50ft/5sec. Or 10ft/sec is the rate per sec of the change of velocity .

This rate gives the final velocity after a specific amount of time...
4 seconds... 40 ft/s....

Gravitational Acceleration: Google search yields only incorrect
analyses

Gravitational Acceleration: Google search yields only incorrect
analyses

Acceleration is a derivative of a derivative. Velocity is, a rate of
distance in time. Acceleration then is, rate of change in the rate of
distance in time. The acceleration of gravity is, the distance change
as a function of time. cm/sec^2.

If an object goes from 0 to 50 ft per second in 5 seconds, it's
acceleration is 50 ft per second / 5 seconds. This simplifies to
50ft/5sec. Or 10ft/sec is the rate per sec of the change of velocity .
This rate gives the final velocity after a specific amount of time...
4 seconds... 40 ft/s....
If the final velocity is 50 ft/s after 5 sec, the average velocity in
this time period for uniform acceleration is 1/2 final velocity.
50/2 = 25 ft/s average velocity. Average velocity x time equals
distance. 25 ft/s x 5 sec = 125 ft, final distance at the end of 5
sec.
What is the rate of acceleration? The rate of the velocity change is 10
ft/s. The average of the velocity change within the
average second is however 1/2 x 10 ft/s or 5 ft/sec^2 . This is the
average rate of the change, of the rate of distance in time,
which gives, the function of time for the distance.
Acceleration of 5 ft/s^2 for 5 seconds = total distance 125 ft, final
velocity 50 ft/s, average velocity 25 ft/s

Gravitaional acceleration 9.8 meters/sec^2
approx. 10 meters / sec^2.
rate of change in velocity(secs)= 20 m/s under gravitational
acceleration
time total distance final velocity average velocity
1sec 10m --- 10 m/s
2 40 40 20
3 90 60 30
4 160 80 40
5 250 100 50
6 360 120 60
7 490 140 70

At the end of 4 secs, for example the final velocity is 80 m/s. It has
traveled 160 m . If the acceleration stopped at this point it
will retain the velocity 80 m/s for the next second traveling 80 m .
This is 10 meters short of the distance it would have traveled under
acceleration or 250 m in 5 seconds. If an object were traveling at 80
m/s and increased it's velocity within one second to travel within that
second 90 m or 10 m greater distance, it would have to accelerate at
uniform rate within that second to 100 m/s in order to attain the
average velocity within that second to travel the extra distance of 10
meters. 80 + 100 / 2 = 90 m/s average velocity for one second = 90 m .
It increases it's velocity by 20 m/s. It accelerates 10 m/s^2.

An object that collides with an identical object in an elastic
collision and exchanges all of it's momentum will decelerate with an
average velocity over the time of transfer, 1/2 it's intital velocity ,
while the other object accelerates with an average velocity during the
time of transfer which is 1/2 it's final velocity.

Kent Deatherage
http://home.earthlink.net/~kdthrge