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We'd like to consider
velocity-dependent resistive

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forces.

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So let's imagine a situation
where we have an object,

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and it's falling in a
gravitational field,

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so we have a
gravitational force.

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And this object is like a
marble falling in a vat of oil.

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And there is a
velocity-dependent resistive

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force that is acting
on this object.

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And let's again choose the
positive direction down.

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And now for objects that are
moving very slowly in a fluid,

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let's make a model for our
velocity-dependent resistive

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force.

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We will assume that
the resistive force is

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proportional to
the velocity, it's

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opposing the
direction of motion,

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and we'll put a
coefficient alpha in front.

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And this is our model for
velocity-dependent resistive

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forces for objects that
are moving very slowly.

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For our case, let's consider the
units of the coefficient alpha.

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So the units of
alpha have the units

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of force divided by velocity.

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So the units of force are-- we
have the units of force divided

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by the units of
velocity, and that

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gives us kilogram meter
per second squared, divided

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by the units of velocity,
which are velocity per second.

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And so we see that the
coefficient of alpha

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has the units of kilogram
per inverse second.

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Now what we'd like to do is
apply Newton's Second Law.

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So let's analyze the forces
and get the equation of motion

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for this object falling
in a viscous medium

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with a resistive force.

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In the j hat direction, we
have the gravitational force.

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And we have our
resistive friction force,

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which we're writing Vy as the
y component of the velocity.

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And that's equal
to m times dVy dt.

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Now notice that
the acceleration--

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we're not treating
as a constant.

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It's a derivative
of the velocity.

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This leads us to our
differential equation, which

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is written as dy dt--
I'll divide through

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by m-- g minus alpha over m, Vy.

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Now this is a linear
and velocity first order

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differential equation with
a constant term in here,

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which is called inhomogeneous
linear first order differential

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equation.

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We shall solve this like we
did with our other equations

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by separation of variables.

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So what we do is we
bring the velocity terms

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to one side and the time
terms to the other side.

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And now we can integrate
both sides directly.

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But before I
integrate this side,

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I'd like to rewrite it
by writing it as dVy--

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and I'd like to pull out
a minus alpha over m.

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And this term here becomes
minus mg over alpha.

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Why did I do this?

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For the integration, it just
makes my natural logarithm

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interval look a
little bit easier.

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And over here, I had dt.

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Now what I want to
do is integrate.

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And we're going to integrate
from some initial velocity,

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which we're going
to take to be 0,

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we're going to let this
object be released from rest,

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and we're going to integrate
to some final speed Vy of t.

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Over here, let's just call our
integration variable t prime.

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And we integrate t prime
from 0 to some time t.

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So this is our variable and the
velocity as a function of time.

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Now doing this
interval-- first off,

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this constant term comes out.

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So we have 1 over
minus alpha over m.

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And the interval is just
a natural log integral

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of Vy minus mg over alpha.

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And we have to divide that
in the lower limit, Vy is 0,

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and we have just
minus mg over alpha.

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And this side is very simply t.

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Now we can rewrite this
equation by bringing the m

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over alpha to the other side.

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And we have log of Vy minus
mg alpha, over minus mg alpha,

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equals minus alpha m t.

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Now recall that e to the log of
any quantity x is equal to x.

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So if I exponentiate
both sides, I

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get Vy minus mg over alpha,
divided by minus mg alpha,

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is equal to the exponential
e to the minus alpha m t.

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Now the last step is simply
to bring this over to there,

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multiply, and do a
little bit of algebra.

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And so I get that the
velocity as a function of y

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is equal to mg alpha times 1
minus e to the minus alpha m t.

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Now there's a few
interesting things here

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that we want to look at.

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Suppose we define
this exponential e

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to the minus alpha m t as equal
to e to the minus t over tau,

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where tau is m over alpha.

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The units of tau, when
we look at the units,

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it has the units
of mass, which is

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kilograms, over the units of
alpha, which we worked out

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before.

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So we have kilograms over
kilogram inverse second.

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That gives us units of seconds.

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And that's what we would expect
because t seconds over seconds

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gives us dimensionless
quantity for the exponential.

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Now the last thing
we'd like to look at

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is let's just see what
happens when we plot this.

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Well, if we're plotting
Vy as a function of t,

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we start off at 0.

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And we're going to reach
some terminal quantity Vy

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terminal, which is when
we have a graph like that.

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And we can see what
that terminal speed

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is when we set t
equal to infinity,

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e to the minus infinity is 0.

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And so this terminal speed is
equal to just this coefficient,

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mg over alpha.

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And that's the terminal speed.

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Now notice, that as a
check for what we did,

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we can come back to our
differential equation,

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and what does
terminal speed mean?

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Notice that when you
get to this limit,

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as t approaches infinity, the
slope of the velocity curve

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is zero.

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And that's really the
statement over here

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that Vy terminal is when the
velocity is no longer changing

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in time.

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You can see it graphically.

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And this is the statement, when
we go back to Newton's Second

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Law, that acceleration is 0.

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So the sum of the forces
at terminal speed is 0,

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the resistive force is equal
to the gravitational force,

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and so from this equation
here, as a check,

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we can see that Vy terminal
equals mg over alpha, which

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checks with our calculation.

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And so we have confidence that
we analyzed this correctly.

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And that gives us our motion.

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One last thing that's
quite interesting.

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Notice that the terminal
speed depends on mass.

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This coefficient alpha
is only a function

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of the properties
of the object--

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the size, the shape,
the geometric properties

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of the object, not
how dense it is.

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So if you have two objects,
like two balls, same radius,

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but one was denser than the
other, then the other mass

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would be heavier and
the terminal speed

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would be greater.

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So what we see is that two
identical objects, one heavier

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than the other,
the heavier object

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reaches a faster terminal speed
than the lighter object, which

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explains a well
observed phenomenon.