Rate of free fall vs Acceleration of free fall

"Gregory L. Hansen" wrote in message
...
In article ,
Donald G. Shead wrote:
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage
over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because
no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.




The "tedious integration" is nothing more than that the distance travelled
is equal the *average* speed multiplied by elapsed time. Suppose you
start with v_i=0 and accelerate to v_f=a*t. How far have you travelled?
Should you use 0? No, you haven't been going 0 the whole time. Should
you use a*t? No, you haven't been going a*t the whole time. Take the
average, x=[(0+a*t)/2]*t, average speed multiplied by elapsed time.

If you have an initial velocity, you start at v_i and finish at v_i+a*t.
The average speed is [v_i + (v_i+a*t)]/2=v_i+a*t/2. Distance traveled
equals average speed multiplied by elapsed time,

x = (v_i + a*t/2)*t

If acceleration is not constant it can be hard to find an average speed
without calculus.

If you use the "rate" above, then rate is not equal to the change in speed
over elapsed time.

That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the time:
It applies not only to free fall but to any constant change in position: So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

acceleration: a = (v_f - v_i) / (t_f - t_i)

rate: r = (v_f - v_i) / 2*(t_f - t_i)

Then you won't need those factors of 1/2 when you find distances because
the 1/2 is already built into the constant, but it's still there. It
certainly won't simplify things when you work problems with non-constant
acceleration, but there's no real problem with finding a solution by
calculus and then replacing every a with 2*r.

It's an unnessary tedious repetitious waste of time is what it really is(;^!
A brilliant mind has better things to do than dinking with the calculus to
find an average value between two known values regardless of how close
together they are.
--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible

"Gregory L. Hansen" wrote in message
...
In article ,
Donald G. Shead wrote:
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage
over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because
no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.




The "tedious integration" is nothing more than that the distance travelled
is equal the *average* speed multiplied by elapsed time. Suppose you
start with v_i=0 and accelerate to v_f=a*t. How far have you travelled?
Should you use 0? No, you haven't been going 0 the whole time. Should
you use a*t? No, you haven't been going a*t the whole time. Take the
average, x=[(0+a*t)/2]*t, average speed multiplied by elapsed time.

If you have an initial velocity, you start at v_i and finish at v_i+a*t.
The average speed is [v_i + (v_i+a*t)]/2=v_i+a*t/2. Distance traveled
equals average speed multiplied by elapsed time,

x = (v_i + a*t/2)*t

If acceleration is not constant it can be hard to find an average speed
without calculus.

If you use the "rate" above, then rate is not equal to the change in speed
over elapsed time.

That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the time:
It applies not only to free fall but to any constant change in position: So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

acceleration: a = (v_f - v_i) / (t_f - t_i)

rate: r = (v_f - v_i) / 2*(t_f - t_i)

Then you won't need those factors of 1/2 when you find distances because
the 1/2 is already built into the constant, but it's still there. It
certainly won't simplify things when you work problems with non-constant
acceleration, but there's no real problem with finding a solution by
calculus and then replacing every a with 2*r.

It's an unnessary tedious repetitious waste of time is what it really is(;^!
A brilliant mind has better things to do than dinking with the calculus to
find an average value between two known values regardless of how close
together they are.
--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible

In article ,
Donald G. Shead wrote:

"Gregory L. Hansen" wrote in message
...
In article ,
Donald G. Shead wrote:
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage
over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because
no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.




The "tedious integration" is nothing more than that the distance travelled
is equal the *average* speed multiplied by elapsed time. Suppose you
start with v_i=0 and accelerate to v_f=a*t. How far have you travelled?
Should you use 0? No, you haven't been going 0 the whole time. Should
you use a*t? No, you haven't been going a*t the whole time. Take the
average, x=[(0+a*t)/2]*t, average speed multiplied by elapsed time.

If you have an initial velocity, you start at v_i and finish at v_i+a*t.
The average speed is [v_i + (v_i+a*t)]/2=v_i+a*t/2. Distance traveled
equals average speed multiplied by elapsed time,

x = (v_i + a*t/2)*t

If acceleration is not constant it can be hard to find an average speed
without calculus.

If you use the "rate" above, then rate is not equal to the change in speed
over elapsed time.

That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the time:
It applies not only to free fall but to any constant change in position: So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

Use whatever works for you.


acceleration: a = (v_f - v_i) / (t_f - t_i)

rate: r = (v_f - v_i) / 2*(t_f - t_i)

Then you won't need those factors of 1/2 when you find distances because
the 1/2 is already built into the constant, but it's still there. It
certainly won't simplify things when you work problems with non-constant
acceleration, but there's no real problem with finding a solution by
calculus and then replacing every a with 2*r.

It's an unnessary tedious repetitious waste of time is what it really is(;^!
A brilliant mind has better things to do than dinking with the calculus to
find an average value between two known values regardless of how close
together they are.

A simple but useful problem involving non-constant acceleration is a mass
bouncing on a spring, a model for all sorts of oscillatory and wave-like
behavior.

x(t) = A*cos(2*pi*f*t)

f is frequency in Hertz.

Can you find the acceleration? I can.

--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible

"Gregory L. Hansen" wrote in message
...
Cut

Use whatever works for you.

Cut

A simple but useful problem involving non-constant acceleration is a mass
bouncing on a spring, a model for all sorts of oscillatory and wave-like
behavior.

x(t) = A*cos(2*pi*f*t)

f is frequency in Hertz.

Can you find the acceleration? I can.

Well I dunno Greg; but I'll keep in touch:
--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible

"Donald G. Shead" wrote in message om...
That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the time:
It applies not only to free fall but to any constant change in position: So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

Since it applies to "any constant change in position" whatever
that means, perhaps you can show me how to use s = at^2/2 to
predict what happens in these two cases:

1. A ball is thrown downward at 5 m/sec. It falls under
the influence of gravity after that.

2. A car is driving east at 50 ft/sec.

- Randy

"Randy Poe" wrote in message
om...
"Donald G. Shead" wrote in message
om...
That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the
time:
It applies not only to free fall but to any constant change in position:
So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

Since it applies to "any constant change in position" whatever
that means, perhaps you can show me how to use s = at^2/2 to
predict what happens in these two cases:

1. A ball is thrown downward at 5 m/sec. It falls under
the influence of gravity after that.

2. A car is driving east at 50 ft/sec.

- Randy

You and your stupid questions(;^)! Be patient and don't think that you've
got me beefalloed:

Galileo's; s = at²/2 applies to "any constant change in position"
whatsoever.

On Wed, 17 Sep 2003 21:35:53 GMT, "Donald G. Shead"
wrote:


"Randy Poe" wrote in message
. com...
"Donald G. Shead" wrote in message
. com...
That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the
time:
It applies not only to free fall but to any constant change in position:
So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

Since it applies to "any constant change in position" whatever
that means, perhaps you can show me how to use s = at^2/2 to
predict what happens in these two cases:

1. A ball is thrown downward at 5 m/sec. It falls under
the influence of gravity after that.

2. A car is driving east at 50 ft/sec.

- Randy

You and your stupid questions(;^)! Be patient and don't think that you've
got me beefalloed:

Galileo's; s = at²/2 applies to "any constant change in position"
whatsoever.

He's got you beefalloed, if that's how you want to characterize it.

If you aren't at a complete loss, then the timer's on, Donald. Show
us your calculating skills, and that you really understand the
formulas you keep spewing out and aren't just copying them. Exactly
how quickly you can answer those two questions?
Gene Nygaard
http://ourworld.compuserve.com/homepages/Gene_Nygaard/

"Donald G. Shead" wrote in message om...
"Randy Poe" wrote in message
om...
"Donald G. Shead" wrote in message
om...
That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the
time:
It applies not only to free fall but to any constant change in position:
So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

Since it applies to "any constant change in position" whatever
that means, perhaps you can show me how to use s = at^2/2 to
predict what happens in these two cases:

1. A ball is thrown downward at 5 m/sec. It falls under
the influence of gravity after that.

2. A car is driving east at 50 ft/sec.

- Randy

You and your stupid questions(;^)! Be patient and don't think that you've
got me beefalloed:

Galileo's; s = at²/2 applies to "any constant change in position"
whatsoever.

Fine. So it applies to my particular two. Show me how.

- Randy

"Randy Poe" wrote in message
om...
"Donald G. Shead" wrote in message
om...
"Randy Poe" wrote in message
om...
"Donald G. Shead" wrote in message
om...
That's the whole point of what I'm saying: Galileo's Rate of free
fall
doesn't concern speed, or changes in velocity; it is a rate of
change in
position; where the change in position varies with the square of the
time:
It applies not only to free fall but to any constant change in
position:
So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

Since it applies to "any constant change in position" whatever
that means, perhaps you can show me how to use s = at^2/2 to
predict what happens in these two cases:

1. A ball is thrown downward at 5 m/sec. It falls under
the influence of gravity after that.

2. A car is driving east at 50 ft/sec.

- Randy

You and your stupid questions(;^)! Be patient and don't think that
you've
got me beefalloed:

Galileo's; s = at²/2 applies to "any constant change in position"
whatsoever.

Fine. So it applies to my particular two. Show me how.

- Randy

Well, I don't have the time or patience to argue with you on such silly data
as you supply Randolf; but the whole equation is: d = vi*t + at²/2 = l + s,
and is for _two "kinds"_ of motion: Inertial motion at a constant velocity
[vi]; according to Newon's first law, and "forced" accelerated motion [a]
according to his second law; which add up to (result in) [d]:

So for your first question; vi = 5 m/sec; then a = 32'/sec²; for whatever
the time [t] is.

For your second stupifying question; I'll assume that the car is going
50'/sec to my right; since I like to draw things on paper, and sometimes sit
facing North, with the paper in front of me. Is that okay with you? Then
what?

"Donald G. Shead" wrote in message
m...

"Randy Poe" wrote in message
om...
"Donald G. Shead" wrote in message
om...
"Randy Poe" wrote in message
om...
"Donald G. Shead" wrote in message
om...
That's the whole point of what I'm saying: Galileo's Rate of free
fall
doesn't concern speed, or changes in velocity; it is a rate of
change in
position; where the change in position varies with the square of
the
time:
It applies not only to free fall but to any constant change in
position:
So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

Since it applies to "any constant change in position" whatever
that means, perhaps you can show me how to use s = at^2/2 to
predict what happens in these two cases:

1. A ball is thrown downward at 5 m/sec. It falls under
the influence of gravity after that.

2. A car is driving east at 50 ft/sec.

- Randy

You and your stupid questions(;^)! Be patient and don't think that
you've
got me beefalloed:

Galileo's; s = at²/2 applies to "any constant change in position"
whatsoever.

Fine. So it applies to my particular two. Show me how.

- Randy

Well, I don't have the time or patience to argue with you on such silly
data
as you supply Randolf; but the whole equation is: d = vi*t + at²/2 = l +
s,
and is for _two "kinds"_ of motion: Inertial motion at a constant velocity
[vi]; according to Newon's first law, and "forced" accelerated motion [a]
according to his second law; which add up to (result in) [d]:

So for your first question; vi = 5 m/sec; then a = 32'/sec²; for whatever
the time [t] is.

So what is the velocity and postion after 10 seconds?


For your second stupifying question; I'll assume that the car is going
50'/sec to my right; since I like to draw things on paper, and sometimes
sit
facing North, with the paper in front of me. Is that okay with you? Then
what?

Same question: what is the velocity and position after 10 seconds? Don't
forget to use 1/2(at^2)!