| Topic: |
Science > Physics |
| User: |
"Don1" |
| Date: |
24 Sep 2005 09:04:37 AM |
| Object: |
The rate of free fall is g/2 |
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
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| User: "KeithK" |
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| Title: Re: The rate of free fall is g/2 |
25 Sep 2005 01:14:07 AM |
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"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance that
the gravity field can be approximated as constant, the acceleration, a(t),
of a body in free fall is given by:
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed time.
Cheers,
KeithK
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
25 Sep 2005 07:01:15 AM |
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KeithK wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance that
the gravity field can be approximated as constant, the acceleration, a(t),
of a body in free fall is given by:
Since when Keith, has acceleration become a(t)? Since Hector was a pup
average acceleration has been a.
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed time.
You must be thinking of bodies falling from sky-high; that take a long
time to fall.
Cheers,
KeithK
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| User: "KeithK" |
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| Title: Re: The rate of free fall is g/2 |
25 Sep 2005 12:30:38 PM |
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"Don1" <dcshead@charter.net> wrote in message
news:1127649674.990528.52600@f14g2000cwb.googlegroups.com...
KeithK wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance
that
the gravity field can be approximated as constant, the acceleration,
a(t),
of a body in free fall is given by:
Since when Keith, has acceleration become a(t)? Since Hector was a pup
average acceleration has been a.
'average acceleration'? what does that have to do with anything? In
general, acceleration can be a function of time, denoted by a(t). Within
the restrictions I cited, a body approximately falls with constant
acceleration g. Thus as time t passes a(t) in this case is approximated by
a(t) = g.
KeithK
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your
g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is
found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed
time.
You must be thinking of bodies falling from sky-high; that take a long
time to fall.
Cheers,
KeithK
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| User: "TomGee" |
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| Title: Re: The rate of free fall is g/2 |
25 Sep 2005 03:03:09 PM |
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Don1 wrote:
KeithK wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance that
the gravity field can be approximated as constant, the acceleration, a(t),
of a body in free fall is given by:
Since when Keith, has acceleration become a(t)? Since Hector was a pup
average acceleration has been a.
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed time.
You must be thinking of bodies falling from sky-high; that take a long
time to fall.
Cheers,
KeithK
To me, you have been rather silly lately, but this is interesting. It
seems few here know what rate means and they define it as other than
what it is.
Rate is an amount of something expressed as a proportion of and in
relation to a whole. If g relates to the gravity at a specific time
(as compared to a range of time), the rate of fall of an object should
be calculable at any point of its travel downward by comparison to it's
initial rate of fall. Why do you divide the gravitational force by 2?
It seems to me that the inertial force of the body offers some
resistance the increase of the rate of acceleration so that must be
considered as well. How about explaining your idea a little more to us
laypersons?
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| User: "Schoenfeld" |
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| Title: Re: The rate of free fall is g/2 |
26 Sep 2005 10:06:15 AM |
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TomGee wrote:
Don1 wrote:
KeithK wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance that
the gravity field can be approximated as constant, the acceleration, a(t),
of a body in free fall is given by:
Since when Keith, has acceleration become a(t)? Since Hector was a pup
average acceleration has been a.
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed time.
You must be thinking of bodies falling from sky-high; that take a long
time to fall.
Cheers,
KeithK
To me, you have been rather silly lately, but this is interesting. It
seems few here know what rate means and they define it as other than
what it is.
Rate is an amount of something expressed as a proportion of and in
relation to a whole. If g relates to the gravity at a specific time
(as compared to a range of time), the rate of fall of an object should
be calculable at any point of its travel downward by comparison to it's
initial rate of fall. Why do you divide the gravitational force by 2?
It seems to me that the inertial force of the body offers some
resistance the increase of the rate of acceleration so that must be
considered as well. How about explaining your idea a little more to us
laypersons?
The 'rate of freefall' can be interpreted more precisely as the 'time
rate of change of freefall' which would be the gravitational jerk. This
quantity is dependent on the velocity of the body.
If g(r) = GM/r^2 then dg/dt = (dg/dr)(dr/dt) = - 2 G M / r^3 (dr/dt).
Here (dr/dt) denotes the component of velocity radial to the M's centre
of gravity. This quantity depends on velocity. For example, the rate of
freefall of a levitating or orbiting body is constant. For a body with
large velocity the jerk is proportionally larger.
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
26 Sep 2005 06:33:41 PM |
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Schoenfeld wrote:
TomGee wrote:
Don1 wrote:
KeithK wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance that
the gravity field can be approximated as constant, the acceleration, a(t),
of a body in free fall is given by:
Since when Keith, has acceleration become a(t)? Since Hector was a pup
average acceleration has been a.
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed time.
You must be thinking of bodies falling from sky-high; that take a long
time to fall.
Cheers,
KeithK
To me, you have been rather silly lately, but this is interesting. It
seems few here know what rate means and they define it as other than
what it is.
Rate is an amount of something expressed as a proportion of and in
relation to a whole. If g relates to the gravity at a specific time
(as compared to a range of time), the rate of fall of an object should
be calculable at any point of its travel downward by comparison to it's
initial rate of fall. Why do you divide the gravitational force by 2?
It seems to me that the inertial force of the body offers some
resistance the increase of the rate of acceleration so that must be
considered as well. How about explaining your idea a little more to us
laypersons?
The 'rate of freefall' can be interpreted more precisely as the 'time
rate of change of freefall' which would be the gravitational jerk. This
quantity is dependent on the velocity of the body.
If g(r) = GM/r^2 then dg/dt = (dg/dr)(dr/dt) = - 2 G M / r^3 (dr/dt).
Here (dr/dt) denotes the component of velocity radial to the M's centre
of gravity. This quantity depends on velocity. For example, the rate of
freefall of a levitating or orbiting body is constant. For a body with
large velocity the jerk is proportionally larger.
_You_ are the jerk: However I just _might_ concede that since s/t
=(vt-vi); is a rate of change in position, g/2 is s/t^2; the rate at
which falling bodies change position per the time they continue to
fall; where the distance fallen in 5 seconds is 16'/sec times 5 sec^2 =
400 feet.
.
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| User: "Schoenfeld" |
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| Title: Re: The rate of free fall is g/2 |
26 Sep 2005 10:12:20 PM |
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Don1 wrote:
Schoenfeld wrote:
TomGee wrote:
Don1 wrote:
KeithK wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance that
the gravity field can be approximated as constant, the acceleration, a(t),
of a body in free fall is given by:
Since when Keith, has acceleration become a(t)? Since Hector was a pup
average acceleration has been a.
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed time.
You must be thinking of bodies falling from sky-high; that take a long
time to fall.
Cheers,
KeithK
To me, you have been rather silly lately, but this is interesting. It
seems few here know what rate means and they define it as other than
what it is.
Rate is an amount of something expressed as a proportion of and in
relation to a whole. If g relates to the gravity at a specific time
(as compared to a range of time), the rate of fall of an object should
be calculable at any point of its travel downward by comparison to it's
initial rate of fall. Why do you divide the gravitational force by 2?
It seems to me that the inertial force of the body offers some
resistance the increase of the rate of acceleration so that must be
considered as well. How about explaining your idea a little more to us
laypersons?
The 'rate of freefall' can be interpreted more precisely as the 'time
rate of change of freefall' which would be the gravitational jerk. This
quantity is dependent on the velocity of the body.
If g(r) = GM/r^2 then dg/dt = (dg/dr)(dr/dt) = - 2 G M / r^3 (dr/dt).
Here (dr/dt) denotes the component of velocity radial to the M's centre
of gravity. This quantity depends on velocity. For example, the rate of
freefall of a levitating or orbiting body is constant. For a body with
large velocity the jerk is proportionally larger.
_You_ are the jerk: However I just _might_ concede that since s/t
=(vt-vi); is a rate of change in position, g/2 is s/t^2; the rate at
which falling bodies change position per the time they continue to
fall; where the distance fallen in 5 seconds is 16'/sec times 5 sec^2 =
400 feet.
v(t) - v(0) is the change in velocity.
[ v(t) - v(0) ] / t is the average acceleration.
[ a(t) - a(0) ] / t is the average jerk.
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| User: "tadchem" |
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| Title: Re: The rate of free fall is g/2 |
27 Sep 2005 07:44:25 AM |
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None of the jerks around here appear to be 'average.'
(TDMMDT)
Tom Davidson
Richmond, VA
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
27 Sep 2005 09:13:20 AM |
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tadchem wrote:
None of the jerks around here appear to be 'average.'
(TDMMDT)
Tom Davidson
Richmond, VA
Oh, I think so, most of us are thick as planks. dense as platinum, and
changable as the weather.
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
27 Sep 2005 06:43:07 AM |
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Schoenfeld wrote:
Don1 wrote:
Schoenfeld wrote:
TomGee wrote:
Don1 wrote:
KeithK wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
Don
Sorry, but you're wrong. Assuming a short enough free-fall distance that
the gravity field can be approximated as constant, the acceleration, a(t),
of a body in free fall is given by:
Since when Keith, has acceleration become a(t)? Since Hector was a pup
average acceleration has been a.
a(t) = g
Therefore it does not fall at a constant rate (velocity) given by your g/2,
since it is (duh) accelerating.
It's rate, or velocity v(t), of freefall is a function of time and is found
by integrating a(t) over time dt:
v(t) = INT(g dt) = g*t
Therefore your statement should have concluded that the rate at which it
freefalls is not constant but is given by g*t where t is the elapsed time.
You must be thinking of bodies falling from sky-high; that take a long
time to fall.
Cheers,
KeithK
To me, you have been rather silly lately, but this is interesting. It
seems few here know what rate means and they define it as other than
what it is.
Rate is an amount of something expressed as a proportion of and in
relation to a whole. If g relates to the gravity at a specific time
(as compared to a range of time), the rate of fall of an object should
be calculable at any point of its travel downward by comparison to it's
initial rate of fall. Why do you divide the gravitational force by 2?
It seems to me that the inertial force of the body offers some
resistance the increase of the rate of acceleration so that must be
considered as well. How about explaining your idea a little more to us
laypersons?
The 'rate of freefall' can be interpreted more precisely as the 'time
rate of change of freefall' which would be the gravitational jerk. This
quantity is dependent on the velocity of the body.
If g(r) = GM/r^2 then dg/dt = (dg/dr)(dr/dt) = - 2 G M / r^3 (dr/dt).
Here (dr/dt) denotes the component of velocity radial to the M's centre
of gravity. This quantity depends on velocity. For example, the rate of
freefall of a levitating or orbiting body is constant. For a body with
large velocity the jerk is proportionally larger.
_You_ are the jerk: However I just _might_ concede that since s/t
=(vt-vi); is a rate of change in position, g/2 is s/t^2; the rate at
which falling bodies change position per the time they continue to
fall; where the distance fallen in 5 seconds is 16'/sec times 5 sec^2 =
400 feet.
v(t) - v(0) is the change in velocity.
[ v(t) - v(0) ] / t is the average acceleration.
Yes it is; the average acceleration is a/2, and is equal to the
distance displaced (s), divided by the time (t) during which it occurs.
To account for initial velocities other than starting from rest, why
dont you use vi, instead of v(0).
[ a(t) - a(0) ] / t is the average jerk.
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| User: "OG" |
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| Title: Re: The rate of free fall is g/2 |
28 Sep 2005 06:04:50 PM |
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"Don1" <dcshead@charter.net> wrote in message
news:1127821387.825794.55830@g14g2000cwa.googlegroups.com...
v(t) - v(0) is the change in velocity.
[ v(t) - v(0) ] / t is the average acceleration.
Yes it is; the average acceleration is a/2, and is equal to the
distance displaced (s), divided by the time (t) during which it occurs.
To account for initial velocities other than starting from rest, why
dont you use vi, instead of v(0).
No Don, the distance displaced (s), divided by the time (t) during which it
occurs gives the average _velocity_, not the average acceleration.
A hint is given by the standard formula v=s/t
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
28 Sep 2005 09:04:25 PM |
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OG wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127821387.825794.55830@g14g2000cwa.googlegroups.com...
v(t) - v(0) is the change in velocity.
[ v(t) - v(0) ] / t is the average acceleration.
Yes it is; the average acceleration is a/2, and is equal to the
distance displaced (s), divided by the time (t) during which it occurs.
To account for initial velocities other than starting from rest, why
dont you use vi, instead of v(0).
No Don, the distance displaced (s), divided by the time (t) during which it
occurs gives the average _velocity_, not the average acceleration.
A hint is given by the standard formula v=s/t
No GO the distance displaced (s), divided by the square of the time (t)
during which it
occurs gives the average acceleration (a/2)!
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| User: "OG" |
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| Title: Re: The rate of free fall is g/2 |
29 Sep 2005 07:00:47 PM |
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"Don1" <dcshead@charter.net> wrote in message
news:1127959465.908745.70060@o13g2000cwo.googlegroups.com...
No Don, the distance displaced (s), divided by the time (t) during which
it
occurs gives the average _velocity_, not the average acceleration.
A hint is given by the standard formula v=s/t
No GO the distance displaced (s), divided by the square of the time (t)
during which it
occurs gives the average acceleration (a/2)!
Assuming you realise that you are only talking here about free fall from
rest (or similar constant acceleration case).
In which case, the formula you are thinking of is
s/t^2 = a/2
but a/2 is "half the acceleration" not "the average acceleration"!
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| User: "Schoenfeld" |
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| Title: Re: The rate of free fall is g/2 |
28 Sep 2005 11:06:32 PM |
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Don1 wrote:
OG wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127821387.825794.55830@g14g2000cwa.googlegroups.com...
v(t) - v(0) is the change in velocity.
[ v(t) - v(0) ] / t is the average acceleration.
Yes it is; the average acceleration is a/2, and is equal to the
distance displaced (s), divided by the time (t) during which it occurs.
To account for initial velocities other than starting from rest, why
dont you use vi, instead of v(0).
No Don, the distance displaced (s), divided by the time (t) during which it
occurs gives the average _velocity_, not the average acceleration.
A hint is given by the standard formula v=s/t
No GO the distance displaced (s), divided by the square of the time (t)
during which it
occurs gives the average acceleration (a/2)!
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
.
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
29 Sep 2005 06:56:45 AM |
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Schoenfeld wrote:
Don1 wrote:
OG wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127821387.825794.55830@g14g2000cwa.googlegroups.com...
v(t) - v(0) is the change in velocity.
[ v(t) - v(0) ] / t is the average acceleration.
Yes it is; the average acceleration is a/2, and is equal to the
distance displaced (s), divided by the time (t) during which it occurs.
To account for initial velocities other than starting from rest, why
dont you use vi, instead of v(0).
No Don, the distance displaced (s), divided by the time (t) during which it
occurs gives the average _velocity_, not the average acceleration.
A hint is given by the standard formula v=s/t
No GO the distance displaced (s), divided by the square of the time (t)
during which it
occurs gives the average acceleration (a/2)!
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)! You GO, are _multiplying_ Boobs velocity by the time during
which it occurs. Bob's moving at a constant speed of 10 m/sec!
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| User: "Randy Poe" |
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| Title: Re: The rate of free fall is g/2 |
29 Sep 2005 07:41:39 AM |
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Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2? How do you average a, a, a, a, ... a together
to get a/2?
You GO, are _multiplying_ Boobs velocity by the time during
which it occurs. Bob's moving at a constant speed of 10 m/sec!
So what is his acceleration during that time? s is 10,
t is 1. You just told us that you should divide s by
t^2 to get a/2. So what is his acceleration by your statement?
- Randy
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
30 Sep 2005 07:49:55 AM |
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Randy Poe wrote:
Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2?
Yes; if a is a constant change in velocity [a=(vt-vi)/t], as in free
fall where the change in velocity is about 32' per second, per second:
Then the average change in velocity is 16' per second, per second
[a/2=16'/sec^2].
How do you average a, a, a, a, ... a together
to get a/2?
I suppose you mean a1, a2, a3, a4 .... Then you have several different
accelerations, or "jerks". and need to find tthe mean average: Actually
you'd have several different a/2's; several different (a/2)t^2's, and
several different problems.
You GO, are _multiplying_ Boobs velocity by the time during
which it occurs. Bob's moving at a constant speed of 10 m/sec!
So what is his acceleration during that time? s is 10,
t is 1. You just told us that you should divide s by
t^2 to get a/2. So what is his acceleration by your statement?
- Randy
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| User: "Randy Poe" |
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| Title: Re: The rate of free fall is g/2 |
30 Sep 2005 08:40:32 AM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2?
Yes; if a is a constant change in velocity [a=(vt-vi)/t], as in free
fall where the change in velocity is about 32' per second, per second:
Then the average change in velocity is 16' per second, per second
[a/2=16'/sec^2].
How do you average a, a, a, a, ... a together
to get a/2?
I suppose you mean a1, a2, a3, a4 ...
No, I mean a, a, a, a,.
Constant acceleration means every time you measure the acceleration
you get the same number.
So suppose the acceleration in the first second is a. And in
the second second is a. And in the third, and the fourth,
and the fifth.
Are you saying if I average these measurements together,
average together a, a, a, a, a, I will get a/2? Does "average
acceleration" mean something different to you than "the average
of the acceleration"?
- Randy
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
30 Sep 2005 12:43:45 PM |
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Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2?
Yes; if a is a constant change in velocity [a=(vt-vi)/t], as in free
fall where the change in velocity is about 32' per second, per second:
Then the average change in velocity is 16' per second, per second
[a/2=16'/sec^2].
How do you average a, a, a, a, ... a together
to get a/2?
I suppose you mean a1, a2, a3, a4 ...
No, I mean a, a, a, a,.
Constant acceleration means every time you measure the acceleration
you get the same number.
So suppose the acceleration in the first second is a. And in
the second second is a. And in the third, and the fourth,
and the fifth.
Are you saying if I average these measurements together,
average together a, a, a, a, a, I will get a/2? Does "average
acceleration" mean something different to you than "the average
of the acceleration"?
- Randy
Wacko
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| User: "Randy Poe" |
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| Title: Re: The rate of free fall is g/2 |
30 Sep 2005 03:18:23 PM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2?
Yes; if a is a constant change in velocity [a=(vt-vi)/t], as in free
fall where the change in velocity is about 32' per second, per second:
Then the average change in velocity is 16' per second, per second
[a/2=16'/sec^2].
How do you average a, a, a, a, ... a together
to get a/2?
I suppose you mean a1, a2, a3, a4 ...
No, I mean a, a, a, a,.
Constant acceleration means every time you measure the acceleration
you get the same number.
So suppose the acceleration in the first second is a. And in
the second second is a. And in the third, and the fourth,
and the fifth.
Are you saying if I average these measurements together,
average together a, a, a, a, a, I will get a/2? Does "average
acceleration" mean something different to you than "the average
of the acceleration"?
Wacko
Which part did you disagree with?
You want to use a/2 as the "average acceleration". I want to
know why that's an appropriate term when the acceleration
is always a.
For instance, in free fall, the velocity after 1 second is
32 ft/sec. After 2 seconds it is 64 ft/sec (a = 32 ft/sec^2).
After 3 seconds it is 96 ft/sec (a = 32 ft/sec^2). After 4
seconds it is 128 ft/sec (a = 32 ft/sec^2).
In what sense is the "average acceleration" of this situation
16 feet per second^2?
- Randy
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
30 Sep 2005 10:05:10 PM |
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Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2?
Yes; if a is a constant change in velocity [a=(vt-vi)/t], as in free
fall where the change in velocity is about 32' per second, per second:
Then the average change in velocity is 16' per second, per second
[a/2=16'/sec^2].
How do you average a, a, a, a, ... a together
to get a/2?
I suppose you mean a1, a2, a3, a4 ...
No, I mean a, a, a, a,.
Constant acceleration means every time you measure the acceleration
you get the same number.
So suppose the acceleration in the first second is a. And in
the second second is a. And in the third, and the fourth,
and the fifth.
Are you saying if I average these measurements together,
average together a, a, a, a, a, I will get a/2? Does "average
acceleration" mean something different to you than "the average
of the acceleration"?
Wacko
Which part did you disagree with?
You want to use a/2 as the "average acceleration". I want to
know why that's an appropriate term when the acceleration
is always a.
For instance, in free fall, the velocity after 1 second is
32 ft/sec. After 2 seconds it is 64 ft/sec (a = 32 ft/sec^2).
After 3 seconds it is 96 ft/sec (a = 32 ft/sec^2). After 4
seconds it is 128 ft/sec (a = 32 ft/sec^2).
In what sense is the "average acceleration" of this situation
16 feet per second^2?
- Randy
You haven't been paying attention have you:
In free fall the distance fallen after 1 second is 16'/sec^2 x 1^2
sec=16'. After 2 seconds it is 16'/sec^2 x 2^2 sec=64'. After 3 seconds
it is 16'/sec^2 x 3^2 sec=144' . . . .
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| User: "bill" |
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| Title: Re: The rate of free fall is g/2 |
01 Oct 2005 12:23:42 AM |
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Consider this:
Given:
v = a*t
s = (a/2)*(t^2). Then
v = 2*s/t
Thus it would seem that the rate of fall is independent of 'a'?
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
01 Oct 2005 07:52:11 AM |
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bill wrote:
Consider this:
Given:
v = a*t
s = (a/2)*(t^2). Then
v = 2*s/t
Thus it would seem that the rate of fall is independent of 'a'?
That's a bum steer. Where'd v=at come from?
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| User: "Randy Poe" |
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| Title: Re: The rate of free fall is g/2 |
03 Oct 2005 09:04:09 AM |
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Don1 wrote:
bill wrote:
Consider this:
Given:
v = a*t
s = (a/2)*(t^2). Then
v = 2*s/t
Thus it would seem that the rate of fall is independent of 'a'?
That's a bum steer. Where'd v=at come from?
It's the velocity as a function of time for constant acceleration.
An acceleration of 32 ft/sec^2 means that the velocity increases
by 32 ft/sec every second. That is expressed in v = at with
a = 32.
- Randy
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
03 Oct 2005 09:58:50 AM |
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Randy Poe wrote:
Don1 wrote:
bill wrote:
Consider this:
Given:
v = a*t
s = (a/2)*(t^2). Then
v = 2*s/t
Thus it would seem that the rate of fall is independent of 'a'?
That's a bum steer. Where'd v=at come from?
It's the velocity as a function of time for constant acceleration.
Velocity is changing; whether the acceleration is constant or not.
An acceleration of 32 ft/sec^2 means that the velocity increases
by 32 ft/sec every second. That is expressed in v = at with
a = 32.
- Randy
Average aceleration is expressed as [(vt-vi)/t=2s/t^2]; so [a/2=s/t^2].
Where s is the displacement distance; caused by a net force.
Don
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| User: "Randy Poe" |
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| Title: Re: The rate of free fall is g/2 |
03 Oct 2005 10:44:19 AM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
bill wrote:
Consider this:
Given:
v = a*t
s = (a/2)*(t^2). Then
v = 2*s/t
Thus it would seem that the rate of fall is independent of 'a'?
That's a bum steer. Where'd v=at come from?
It's the velocity as a function of time for constant acceleration.
Velocity is changing; whether the acceleration is constant or not.
Every one of the equations you usually write is valid
only for constant acceleration. Most require additional
assumptions as well.
An acceleration of 32 ft/sec^2 means that the velocity increases
by 32 ft/sec every second. That is expressed in v = at with
a = 32.
Average aceleration is expressed as [(vt-vi)/t=2s/t^2]; so [a/2=s/t^2].
Where s is the displacement distance; caused by a net force.
Average acceleration can be expressed as (vt-vi)/t. That is
true in general. If you use that, you will find a value of
32 ft/sec^2 for gravity.
Average accleration can be expressed as 2s/t^2 if the
acceleration is constant, but not in general.
In either case, note that you just wrote that average
acceleration is equal to 2s/t^2, and not s/t^2.
- Randy
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| User: "bill" |
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| Title: Re: The rate of free fall is g/2 |
01 Oct 2005 08:08:49 PM |
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How would you calculate the velocity?
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| User: "Eric Gisse" |
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| Title: Re: The rate of free fall is g/2 |
30 Sep 2005 11:50:01 PM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2?
Yes; if a is a constant change in velocity [a=(vt-vi)/t], as in free
fall where the change in velocity is about 32' per second, per second:
Then the average change in velocity is 16' per second, per second
[a/2=16'/sec^2].
How do you average a, a, a, a, ... a together
to get a/2?
I suppose you mean a1, a2, a3, a4 ...
No, I mean a, a, a, a,.
Constant acceleration means every time you measure the acceleration
you get the same number.
So suppose the acceleration in the first second is a. And in
the second second is a. And in the third, and the fourth,
and the fifth.
Are you saying if I average these measurements together,
average together a, a, a, a, a, I will get a/2? Does "average
acceleration" mean something different to you than "the average
of the acceleration"?
Wacko
Which part did you disagree with?
You want to use a/2 as the "average acceleration". I want to
know why that's an appropriate term when the acceleration
is always a.
For instance, in free fall, the velocity after 1 second is
32 ft/sec. After 2 seconds it is 64 ft/sec (a = 32 ft/sec^2).
After 3 seconds it is 96 ft/sec (a = 32 ft/sec^2). After 4
seconds it is 128 ft/sec (a = 32 ft/sec^2).
In what sense is the "average acceleration" of this situation
16 feet per second^2?
- Randy
You haven't been paying attention have you:
In free fall the distance fallen after 1 second is 16'/sec^2 x 1^2
sec=16'. After 2 seconds it is 16'/sec^2 x 2^2 sec=64'. After 3 seconds
it is 16'/sec^2 x 3^2 sec=144' . . . .
What is your ***** point?
I never thought I would see a physical example of an infinite loop
until I saw you.
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
01 Oct 2005 07:48:50 AM |
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Eric Gisse wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Schoenfeld wrote:
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)!
Are you saying that if the acceleration at all time is a, then
the average is a/2?
Yes; if a is a constant change in velocity [a=(vt-vi)/t], as in free
fall where the change in velocity is about 32' per second, per second:
Then the average change in velocity is 16' per second, per second
[a/2=16'/sec^2].
How do you average a, a, a, a, ... a together
to get a/2?
I suppose you mean a1, a2, a3, a4 ...
No, I mean a, a, a, a,.
Constant acceleration means every time you measure the acceleration
you get the same number.
So suppose the acceleration in the first second is a. And in
the second second is a. And in the third, and the fourth,
and the fifth.
Are you saying if I average these measurements together,
average together a, a, a, a, a, I will get a/2? Does "average
acceleration" mean something different to you than "the average
of the acceleration"?
Wacko
Which part did you disagree with?
You want to use a/2 as the "average acceleration". I want to
know why that's an appropriate term when the acceleration
is always a.
For instance, in free fall, the velocity after 1 second is
32 ft/sec. After 2 seconds it is 64 ft/sec (a = 32 ft/sec^2).
After 3 seconds it is 96 ft/sec (a = 32 ft/sec^2). After 4
seconds it is 128 ft/sec (a = 32 ft/sec^2).
In what sense is the "average acceleration" of this situation
16 feet per second^2?
- Randy
You haven't been paying attention have you:
In free fall the distance fallen after 1 second is 16'/sec^2 x 1^2
sec=16'. After 2 seconds it is 16'/sec^2 x 2^2 sec=64'. After 3 seconds
it is 16'/sec^2 x 3^2 sec=144' . . . .
What is your ***** point?
I never thought I would see a physical example of an infinite loop
until I saw you.
You're loopier
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| User: "Don1" |
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| Title: Re: The rate of free fall is g/2 |
29 Sep 2005 07:54:19 AM |
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Schoenfeld wrote:
Don1 wrote:
OG wrote:
"Don1" <dcshead@charter.net> wrote in message
news:1127821387.825794.55830@g14g2000cwa.googlegroups.com...
v(t) - v(0) is the change in velocity.
[ v(t) - v(0) ] / t is the average acceleration.
Yes it is; the average acceleration is a/2, and is equal to the
distance displaced (s), divided by the time (t) during which it occurs.
To account for initial velocities other than starting from rest, why
dont you use vi, instead of v(0).
No Don, the distance displaced (s), divided by the time (t) during which it
occurs gives the average _velocity_, not the average acceleration.
A hint is given by the standard formula v=s/t
No GO the distance displaced (s), divided by the square of the time (t)
during which it
occurs gives the average acceleration (a/2)!
Say Bob moves at 1 meter per second.
After 10 seconds, Bob moved 10 meters.
What was Bob's average acceleration?
Listen to me GO: the distance displaced (s), _divided_ by the square of
the time (t) during which it occurs gives the average acceleration
(a/2)! You GO, are _multiplying_ Boobs velocity by the time during
which it occurs. Bob's moving at a constant speed of 10 m/sec!
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| User: "OG" |
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| Title: Re: The rate of free fall is g/2 |
24 Sep 2005 05:36:26 PM |
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"Don1" <dcshead@charter.net> wrote in message
news:1127570676.990512.113690@g14g2000cwa.googlegroups.com...
Whereas the acceleration of free fall is g; the rate of free fall is
g/2.
What do you mean by the 'Rate' of free fall?
the acceleration ?
the speed ?
one of the skills of a scientist is to produce a definiton of a phenomenon
that allows a base for further thought.
In failing to define your terms and you are consistently struggling to keep
a clear picture of what you are discussing.
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