| Topic: |
Science > Physics |
| User: |
"Don1" |
| Date: |
17 Sep 2005 07:56:11 AM |
| Object: |
Formula #4: s=(vi)t + (a/2)t^2 |
Using Formula #4: s=(vi)t + (a/2)t^2:
1) A truck starting from rest acquires a speed of 30 mi/hr in 20
seconds. What is the acceleration, and how far does the truck travel?
Using a=(vt-vi)/t; a = (44'/sec - 0)/t = 2.2'/sec^2.
Using s=(vi)t + (a/2)t^2; s = 0 + [(2.2'/sec^2)/2] x 20 sec^2 = 440
feet.
2) A truck starting from 15 mi/hr acquires a speed of 30 mi/hr in 20
seconds. What is the acceleration, and how far does the truck travel?
Using a=(vt-vi)/t; a = (44'/sec - 22'/sec^2)/t = 2.2'/sec^2.
Using s=(vi)t + (a/2)t^2; s = (22'/sec)20 + [(2.2'/sec^2)/2] x 20 sec^2
= 880 feet.
There is a substantial increase in how far the truck travels when
starting from rest and when starting from 15 mi/hr.
Don
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| User: "Sam Wormley" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 08:01:19 AM |
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Galileo's formula for distance fallen is s = 1/2 a t^2 and
is applicable for zero starting distance and zero starting
velocity.
The more general form starts with Newton's Second Law which
can calculate distance given any starting position and
velocity.
Newton's Second Law
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
"A force F acting on a body gives it an acceleration a which is
in the direction of the force and has magnitude inversely
proportional to the mass m of the body: F = ma"
F = ma is a differential equation and therein lies its power.
Case 1. The force is constant (as Galileo assumed):
Assuming that the mass remains constant, we have constant
acceleration: F/m = dv/dt = a = constant
direct integration (with respect to t) gives formulas such as:
v - v_o = at
v = at + v_o
A second integration results in:
x - x_o = 1/2 at^2 + v_ot
x = 1/2 at^2 + v_ot + x_o
You, Shead, can calculate distance given any starting position and
velocity with s = 1/2 at^2 + v_ot + s_o .
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| User: "Don1" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 08:45:33 AM |
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Sam Wormley wrote:
Galileo's formula for distance fallen is s = 1/2 a t^2 and
is applicable for zero starting distance and zero starting
velocity.
The more general form starts with Newton's Second Law which
can calculate distance given any starting position and
velocity.
Newton's Second Law
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
"A force F acting on a body gives it an acceleration a which is
in the direction of the force and has magnitude inversely
proportional to the mass m of the body: F = ma"
F = ma is a differential equation and therein lies its power.
Case 1. The force is constant (as Galileo assumed):
Assuming that the mass remains constant, we have constant
acceleration: F/m = dv/dt = a = constant
direct integration (with respect to t) gives formulas such as:
v - v_o = at
v = at + v_o
A second integration results in:
x - x_o = 1/2 at^2 + v_ot
x = 1/2 at^2 + v_ot + x_o
You, Shead, can calculate distance given any starting position and
velocity with s = 1/2 at^2 + v_ot + s_o .
It's easier with s=(vi)t + (a/2)t^2, and you don't even need to know
the starting position.
Don
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| User: "Herman Trivilino" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 01:55:45 PM |
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"Don1" <dcshead@charter.net> wrote ...
It's easier with s=(vi)t + (a/2)t^2,
It is easier. And obviously you find it more satisfactory. It doesn't give
results, though, that match what's observed. If you make a graph of s
versus t using your equation, and then make a graph of the observed values
of s and t, the graph doesn't match observed values *unless* the object's
motion is in a straight line with constant acceleration.
and you don't even need to know the starting position.
No, you don't. Unless, that is, you're interested in knowing the final
position. The displacement s is, by definition, the difference between the
initial and final positions. If you don't know the initial position, you
can't use s to find the final position.
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| User: "Don1" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 02:14:33 PM |
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Herman Trivilino wrote:
"Don1" <dcshead@charter.net> wrote ...
It's easier with s=(vi)t + (a/2)t^2,
It is easier. And obviously you find it more satisfactory. It doesn't give
results, though, that match what's observed. If you make a graph of s
versus t using your equation, and then make a graph of the observed values
of s and t, the graph doesn't match observed values *unless* the object's
motion is in a straight line with constant acceleration.
and you don't even need to know the starting position.
No, you don't. Unless, that is, you're interested in knowing the final
position. The displacement s is, by definition, the difference between the
initial and final positions. If you don't know the initial position, you
can't use s to find the final position.
Oh darn!
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| User: "Herman Trivilino" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 04:10:07 PM |
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"Don1" <dcshead@charter.net> wrote ...
Herman Trivilino wrote:
"Don1" <dcshead@charter.net> wrote ...
It's easier with s=(vi)t + (a/2)t^2,
It is easier. And obviously you find it more satisfactory. It doesn't
give
results, though, that match what's observed. If you make a graph of s
versus t using your equation, and then make a graph of the observed
values
of s and t, the graph doesn't match observed values *unless* the object's
motion is in a straight line with constant acceleration.
and you don't even need to know the starting position.
No, you don't. Unless, that is, you're interested in knowing the final
position. The displacement s is, by definition, the difference between
the
initial and final positions. If you don't know the initial position, you
can't use s to find the final position.
Oh darn!
You said it.
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| User: "Schoenfeld" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 11:45:49 AM |
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Don1 wrote:
It's easier with s=(vi)t + (a/2)t^2, and you don't even need to know
the starting position.
Your equation assumes that s(0) = 0. In general terms, you cannot
assume that the acceleration is constant, or that the jerk is constant,
or that the snap is constant, or that the crackle is constant, or that
the pop is constant, hence
s(t) = SUM(n=0)^+INF s^n(t) t^n / n!
which is simply the called the Mclaurin series of 's' (the McLaurin
series is the Taylor series about 0).
Don
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| User: "odin" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 12:02:19 PM |
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Your equation assumes that s(0) = 0. In general terms, you cannot
assume that the acceleration is constant, or that the jerk is constant,
or that the snap is constant, or that the crackle is constant, or that
the pop is constant, hence
s(t) = SUM(n=0)^+INF s^n(t) t^n / n!
The only jerk that is constant seems to be Don1.
BTW, we know that g = 9.8 at sea level. We know that g is a function of s.
Does anyone know what the jerk of g is at sea level?
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| User: "Steve Ralph" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 01:02:29 PM |
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"odin" <ragnarok@yahoo.com> wrote in message
news:TNudnVWwJvCB0bHeRVn-vA@whidbeytel.com...
Your equation assumes that s(0) = 0. In general terms, you cannot
assume that the acceleration is constant, or that the jerk is constant,
or that the snap is constant, or that the crackle is constant, or that
the pop is constant, hence
s(t) = SUM(n=0)^+INF s^n(t) t^n / n!
The only jerk that is constant seems to be Don1.
BTW, we know that g = 9.8 at sea level. We know that g is a function of s.
Does anyone know what the jerk of g is at sea level?
A slug?
sr
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| User: "" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
18 Sep 2005 06:49:04 AM |
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odin wrote:
Your equation assumes that s(0) = 0. In general terms, you cannot
assume that the acceleration is constant, or that the jerk is constant,
or that the snap is constant, or that the crackle is constant, or that
the pop is constant, hence
s(t) = SUM(n=0)^+INF s^n(t) t^n / n!
The only jerk that is constant seems to be Don1.
BTW, we know that g = 9.8 at sea level. We know that g is a function of s.
Does anyone know what the jerk of g is at sea level?
The gravitational jerk varies with the velocity of a body. The
gravitational jerk is the time-deriveative of g.
g(s) = - G M / s^2 {note: at sea level g = -9.8 m/s^2 }
Since s is strictly a function of time t, using the chain rule:
dg/dt = (dg/ds) (ds/dt)
= 2 G M / s^3 (ds/dt)
Here ds/dt denotes the component of velocity radial to the
gravitational source center.
Taking ds/dt = 1 m/s , you can compute a "unit jerk field"
jerk(s) = 2 GM / s^3
which at sea level gives approx 3.073E-6 m/s^3.
note: the "unit jerk" has opposite sign to g.
note2: jerk(s) is numerically equivalent to dg/ds.
[my original reply got lost in Google's systems, if it appears excuse
the double reply]
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| User: "Sam Wormley" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 08:50:34 AM |
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Don1 wrote:
Sam Wormley wrote:
Galileo's formula for distance fallen is s = 1/2 a t^2 and
is applicable for zero starting distance and zero starting
velocity.
The more general form starts with Newton's Second Law which
can calculate distance given any starting position and
velocity.
Newton's Second Law
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
"A force F acting on a body gives it an acceleration a which is
in the direction of the force and has magnitude inversely
proportional to the mass m of the body: F = ma"
F = ma is a differential equation and therein lies its power.
Case 1. The force is constant (as Galileo assumed):
Assuming that the mass remains constant, we have constant
acceleration: F/m = dv/dt = a = constant
direct integration (with respect to t) gives formulas such as:
v - v_o = at
v = at + v_o
A second integration results in:
x - x_o = 1/2 at^2 + v_ot
x = 1/2 at^2 + v_ot + x_o
You, Shead, can calculate distance given any starting position and
velocity with s = 1/2 at^2 + v_ot + s_o .
It's easier with s=(vi)t + (a/2)t^2, and you don't even need to know
the starting position.
Don
You sound like the Bush Administration... "It's easier with..."
In science, one should do science not what's "easier"!
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| User: "Don1" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 11:07:01 AM |
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Sam Wormley wrote:
Don1 wrote:
Sam Wormley wrote:
Galileo's formula for distance fallen is s = 1/2 a t^2 and
is applicable for zero starting distance and zero starting
velocity.
The more general form starts with Newton's Second Law which
can calculate distance given any starting position and
velocity.
Newton's Second Law
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
"A force F acting on a body gives it an acceleration a which is
in the direction of the force and has magnitude inversely
proportional to the mass m of the body: F = ma"
F = ma is a differential equation and therein lies its power.
Case 1. The force is constant (as Galileo assumed):
Assuming that the mass remains constant, we have constant
acceleration: F/m = dv/dt = a = constant
direct integration (with respect to t) gives formulas such as:
v - v_o = at
v = at + v_o
A second integration results in:
x - x_o = 1/2 at^2 + v_ot
x = 1/2 at^2 + v_ot + x_o
You, Shead, can calculate distance given any starting position and
velocity with s = 1/2 at^2 + v_ot + s_o .
It's easier with s=(vi)t + (a/2)t^2, and you don't even need to know
the starting position.
Don
You sound like the Bush Administration... "It's easier with..."
In science, one should do science not what's "easier"!
Hey that's my line; regarding the metric system.
Don
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| User: "Douglas Eagleson" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 11:01:22 AM |
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Sam Wormley wrote:
Don1 wrote:
Sam Wormley wrote:
Galileo's formula for distance fallen is s = 1/2 a t^2 and
is applicable for zero starting distance and zero starting
velocity.
The more general form starts with Newton's Second Law which
can calculate distance given any starting position and
velocity.
Newton's Second Law
http://scienceworld.wolfram.com/physics/NewtonsSecondLaw.html
"A force F acting on a body gives it an acceleration a which is
in the direction of the force and has magnitude inversely
proportional to the mass m of the body: F = ma"
F = ma is a differential equation and therein lies its power.
Case 1. The force is constant (as Galileo assumed):
Assuming that the mass remains constant, we have constant
acceleration: F/m = dv/dt = a = constant
direct integration (with respect to t) gives formulas such as:
v - v_o = at
v = at + v_o
A second integration results in:
x - x_o = 1/2 at^2 + v_ot
x = 1/2 at^2 + v_ot + x_o
You, Shead, can calculate distance given any starting position and
velocity with s = 1/2 at^2 + v_ot + s_o .
It's easier with s=(vi)t + (a/2)t^2, and you don't even need to know
the starting position.
Don
You sound like the Bush Administration... "It's easier with..."
In science, one should do science not what's "easier"!
You might get the differential and then understand the physics. A
function that works in a nonfalsified fashion is a truely quizical
thing.
And mathematic is real subtle making the given function a certain
solution to the general case.
A body is at freefall with the initial velocity due to the independent
cause.
Maybe the rocket engine or maybe the previous freefall.
And the differential is funny like this because the initial state is
the variable in its own solution sometimes.
A form of theory to be remembered is the example that assists in
mathematical theory remebering. A velocity in the second order is the
usage. And the function to derive this one is the only question.
A good theorist would start worrying because acceleration is general.
And if it works for gravity then the initial state of the body is
second not first order related.
And in theory second order is the question. Physical interpretation of
force applied must conform.
And your science can not conform. So you ignore the science and
denigrate.
Douglas Eagleson
Gaitherbsurg, MS USA
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| User: "odin" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 11:25:02 AM |
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[stuff snipped]
Ummm.... sorry I was not paying close attention... could you run that by me
again?
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| User: "Douglas Eagleson" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 03:14:09 PM |
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odin wrote:
[stuff snipped]
Ummm.... sorry I was not paying close attention... could you run that by me
again?
A second order theory is a simple relation. And to have force exist in
the equation's form a fractional change is caused by the existence.
While the exact value is caused by the example force.
A paradox form of theory because I can actually resolve the falsifiable
dilemma. A force to cause a change is a simple force following the
theory of all kinds.
And to allow the force effect without measurable example is reality in
theory. A true existence of force develops. And the abstract force of
newton does not address the cause of state.
A force simply needs to have caused the previous acceleration to vi in
order to make it falsifiable.
It is an independent effect on the body and may mathematically be so
written. A valid theory of force is falsified the second the theory is
state in logical analysis.
And the force of the solid state is hereby proven as the correct
gravitational theory. A true state of the universe is a force.
Douglas Eagleson
Gaithersburg, MD USA
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| User: "odin" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 03:20:28 PM |
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odin wrote:
[stuff snipped]
Ummm.... sorry I was not paying close attention... could you run that by
me
again?
[even more wild eyed stuff snipped]
Gasp! OK. Sorry... thanks for sharing... but that will be enough for now....
thanks again... bye...
.
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| User: "Douglas Eagleson" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
18 Sep 2005 01:36:54 PM |
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odin wrote:
odin wrote:
[stuff snipped]
Ummm.... sorry I was not paying close attention... could you run that by
me
again?
[even more wild eyed stuff snipped]
Gasp! OK. Sorry... thanks for sharing... but that will be enough for now....
thanks again... bye...
Run to school and learn even predicate diaper training reading.
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| User: "Sam Wormley" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 08:48:10 AM |
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Don1 wrote:
Sam Wormley wrote:
You, Shead, can calculate distance given any starting position and
velocity with s = 1/2 at^2 + v_ot + s_o .
It's easier with s=(vi)t + (a/2)t^2, and you don't even need to know
the starting position.
Don
No it's not.
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| User: "Herman Trivilino" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 10:32:17 AM |
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"Don1" <dcshead@charter.net> wrote ...
1) A truck starting from rest acquires a speed of 30 mi/hr in 20
seconds. What is the acceleration, and how far does the truck travel?
Using a=(vt-vi)/t; a = (44'/sec - 0)/t = 2.2'/sec^2.
2.2 ft/s² is the *average* acceleration.
Using s=(vi)t + (a/2)t^2; s = 0 + [(2.2'/sec^2)/2] x 20 sec^2 = 440
feet.
440 ft is the distance traveled *if* the truck's acceleration is constant.
That's *not* a successful model of the way a truck actually behaves.
Consider the 1999 Plymouth Prowler, tested by Motorweek. According to their
observations this car travelled a distance 0.25 miles, accelerating from
rest to a final speed of 98 mi/h in a time of 14.2 seconds.
Using your technique, we calculate the acceleration.
a=(vt-vi)/t; a = (143.7 ft/s - 0)/t = 10.1 ft/s².
Note that this is the *average* acceleration. Continuing to apply your
technique ...
s=(vi)t + (a/2)t^2; s = 0 + [(10.1 ft/s²)/2] x (14.2 s)² = 1020 ft..
Consider the fact that 0.25 miles is 1320 ft.
Evidently, the Plymouth Prowler didn't speed up by 10.1 ft/s during *each*
second of its 14.2 second journey. During the first seconds it sped up by
more than 10.1 ft/s each second, and during the latter seconds less than
10.1 ft/s each second. The *average* amount sped up per second was 10.1
ft/s.
The result is that during this 14.2 second interval it was able to travel
the 1320 ft length of the quarter-mile track. If it were racing against
another car that did have a constant acceleration, the other car would have
lost the race by a distance of 300 ft.
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| User: "Don1" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 11:02:11 AM |
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Herman Trivilino wrote:
"Don1" <dcshead@charter.net> wrote ...
1) A truck starting from rest acquires a speed of 30 mi/hr in 20
seconds. What is the acceleration, and how far does the truck travel?
Using a=(vt-vi)/t; a = (44'/sec - 0)/t = 2.2'/sec^2.
2.2 ft/s2 is the *average* acceleration.
Using s=(vi)t + (a/2)t^2; s = 0 + [(2.2'/sec^2)/2] x 20 sec^2 = 440
feet.
440 ft is the distance traveled *if* the truck's acceleration is constant.
That's *not* a successful model of the way a truck actually behaves.
Consider the 1999 Plymouth Prowler, tested by Motorweek. According to their
observations this car travelled a distance 0.25 miles, accelerating from
rest to a final speed of 98 mi/h in a time of 14.2 seconds.
Using your technique, we calculate the acceleration.
a=(vt-vi)/t; a = (143.7 ft/s - 0)/t = 10.1 ft/s2.
Note that this is the *average* acceleration. Continuing to apply your
technique ...
s=(vi)t + (a/2)t^2; s = 0 + [(10.1 ft/s2)/2] x (14.2 s)2 = 1020 ft..
Consider the fact that 0.25 miles is 1320 ft.
Evidently, the Plymouth Prowler didn't speed up by 10.1 ft/s during *each*
second of its 14.2 second journey. During the first seconds it sped up by
more than 10.1 ft/s each second, and during the latter seconds less than
10.1 ft/s each second. The *average* amount sped up per second was 10.1
ft/s.
The result is that during this 14.2 second interval it was able to travel
the 1320 ft length of the quarter-mile track. If it were racing against
another car that did have a constant acceleration, the other car would have
lost the race by a distance of 300 ft.
Interesting: Must be the difference between how they "burn rubber".
Don
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| User: "Herman Trivilino" |
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| Title: Re: Formula #4: s=(vi)t + (a/2)t^2 |
17 Sep 2005 01:50:03 PM |
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"Don1" <dcshead@charter.net> wrote ...
Evidently, the Plymouth Prowler didn't speed up by 10.1 ft/s during
*each*
second of its 14.2 second journey. During the first seconds it sped up
by
more than 10.1 ft/s each second, and during the latter seconds less than
10.1 ft/s each second. The *average* amount sped up per second was 10.1
ft/s.
The result is that during this 14.2 second interval it was able to travel
the 1320 ft length of the quarter-mile track. If it were racing against
another car that did have a constant acceleration [of 10.1 ft/s²], the
other car would have lost the race by a distance of 300 ft.
Interesting: Must be the difference between how they "burn rubber".
I'm glad you find it interesting. Does that mean you see it?
If you see it, then it behooves you to see that your constant-acceleration
model doesn't match what's observed. It may satisfy you in some way, but it
doesn't match what's observed. That means it doesn't make for good physics.
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