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

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/phys...SecondLaw.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 .

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/phys...SecondLaw.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

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.

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/phys...SecondLaw.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"!

"Don1" 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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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/phys...SecondLaw.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

Herman Trivilino wrote:
"Don1" 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

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/phys...SecondLaw.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

[stuff snipped]

Ummm.... sorry I was not paying close attention... could you run that by me
again?