What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?
Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

Don

What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?

Rate of displacement, is just called velocity which is a function of time
v(t).
Impulse is the integral of force with respect to time.
If force is constant (keeping it simple for Don1), impulse is I = F *
(t1-t0).

Let t = t1-t0
So I = F*t
And F=M*A
So I = M*A*t
With constant force and mass, we have constant acceleration. Assumimg v(0) =
0, we have v(t) = A*t
So I = M*v(t)

You asked what is the ratio of impulse I to velocity v(t).
So I/v(t) = (M*v(t))/v(t) = M.

The ratio you are looking for is just the mass M.

Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

No.

What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?

Rate of displacement, is just called velocity which is a function of time
v(t).
Impulse is the integral of force with respect to time.

To make it simpler, just think of impulse as being "change in momentum" (by
definition). So it has the same units, dimensions, and meaning as momentum.
You know that momentum is mass times velocity (i hope), or p=m*v. You are
asking what is the ratio of momentum to velocity, or p/v. Even Don1 should
see that this ratio is just mass m. Or maybe not...

"Don1" wrote ...

What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?
Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

Impulse equals change in momentum. Impulse equals the product of force and
time *only* when the force is constant. Momentum equals mass times
velocity. Change in momentum equals final momentum minus initial momentum.

Did you loose your introductory physics book?

Send me your address and I'll gladly mail several to you, at my own expense.



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Don1 wrote:
What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?
Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

Don


See: http://scienceworld.wolfram.com/physics/Impulse.html

Herman Trivilino wrote:
"Don1" wrote ...

What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?
Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

Impulse equals change in momentum. Impulse equals the product of force and
time *only* when the force is constant. Momentum equals mass times
velocity. Change in momentum equals final momentum minus initial momentum.

Did you loose your introductory physics book?

I think you lost it too. The impulse that causes a change in momentum
delta(p) in time delta(t) is equal to the average force over that time
interval times delta(t). Force need not to be constant, otherwise
Newton's 2nd law would not hold for arbitrary forcing functions, like
for instance a gravitational potential field that is a function of
position r.

Mike





Send me your address and I'll gladly mail several to you, at my own expense.



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Sam Wormley wrote:
Don1 wrote:
What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?
Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

Don


See: http://scienceworld.wolfram.com/physics/Impulse.html

Oh please Sammy; NOTHING happens instantaneously: Everything takes
time.

Don

Don1 wrote:
Sam Wormley wrote:

Don1 wrote:

What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?
Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

Don


See: http://scienceworld.wolfram.com/physics/Impulse.html


Oh please Sammy; NOTHING happens instantaneously: Everything takes
time.

Don


You probably missed (on purpose?) the integration over time 0-tau!

odin wrote:
What's the ratio of an impulse (ft), to the rate of displacement (s/t)
that it causes?

Rate of displacement, is just called velocity which is a function of time
v(t).
Impulse is the integral of force with respect to time.
If force is constant (keeping it simple for Don1), impulse is I = F *
(t1-t0).

Let t = t1-t0
So I = F*t
And F=M*A
So I = M*A*t
With constant force and mass, we have constant acceleration. Assumimg v(0) =
0, we have v(t) = A*t
So I = M*v(t)

You asked what is the ratio of impulse I to velocity v(t).
So I/v(t) = (M*v(t))/v(t) = M.

The ratio you are looking for is just the mass M.

You got it: The ratio ft/(s/t) can be more concisely as m=ft^2/s; which
is also equal to 2w/g.

Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

No.

First you say it does, and now you say it doesn't. Make up my mind will
you.

Don

You asked what is the ratio of impulse I to velocity v(t).
So I/v(t) = (M*v(t))/v(t) = M.

The ratio you are looking for is just the mass M.

You got it: The ratio ft/(s/t) can be more concisely as m=ft^2/s; which
is also equal to 2w/g.

Is the impulse (ft) numerically equal to the rate of displacement (s/t)
that it causes?

No.

First you say it does, and now you say it doesn't. Make up my mind will
you.

Impulse is not, in general, numerically equal to the rate of displacement.
Assuming an initial velocity of zero, they are proportional to eachother.
The proprotionality is the mass involved. Given a particular system of
units, you can pick the mass carefully to make the impulse numerically equal
to the rate of displacement. In that very specific case you would be
correct. But for the genrality that you framed your question in, where the
initial velocity and the mass involved are arbitrary, the answer to your
question is nope. Note that impulse is a delta, if that helps your
understanding a bit as well.