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Now we're going to analyze
a more complicated example

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of drag forces, where we
have an object falling

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in a gravitational
field with gravity.

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We have a resistive force.

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And this is an object in air.

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And so our model will be
for the resistive force

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that it's going to be
proportional to the velocity

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squared.

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Now to get its direction right,
opposing the motion, j hat.

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So this will turn out to be
a more complicated analysis.

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But first again, let's
think about the units

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of our coefficient beta.

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It has the units of force
divided by velocity squared.

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So we write the units of
force, kilogram meters

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per second squared,
and the units

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of velocity squared, meter
squared per second squared.

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And so we see that we have
units of kilogram per meter

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for our coefficient beta.

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Now, we'll apply as usual
Newton's second law,

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F equals m a, to get
our equation of motion.

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We're looking at
the j direction.

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And our forces are
gravity minus the velocity

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squared resistive drag force.

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And that's equal to the
derivative of the velocity

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dv/dt.

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In this example, it's a
one dimensional motion.

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So I'm dropping any mention
of y direction for simplicity.

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Now we can rewrite
this equation as dv/dt.

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Let's divide through by m is
g minus beta over m v square.

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And this is a linear-- it's
a first order differential

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equation, dv/dt.

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It's a non-linear equation
because the velocity

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term appears here as a square
and there's a constant term.

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But we can still
apply our technique

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of separation of variables.

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And so when we write
this equation as dv/dt

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we'll separate out dv's and
t's, so we have d d times

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g minus beta over and m
v squared is equal to dt.

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Now, I'm going to do two
things just to clean this up

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for algebra a little bit later.

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I'm going to multiply both
sides by a minus sign.

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And I'm going to pull the g out.

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So I have 1 minus beta
over m g v squared.

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And on the other
side, I have minus dt.

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And now I can get this
equation in the form

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that I'd like to
integrate, which

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is minus dv times 1 minus
beta over m g v squared equals

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minus gdt.

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Now, the trouble here is
this integral is a little bit

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complicated.

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So I'd like to make
a change of variable.

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And my change of
variable will be

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u equals the square root of beta
m g times v. And that implies

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that du is beta m g dv.

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And the limits, if we
start our object at rest,

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so if u0 equals 0, then
because v0 equal 0,

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that's our first limit.

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Now we have to be a
little bit careful

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because if we drop this
object at rest, initially

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it will be moving very slowly.

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And so our resistive model
doesn't actually apply.

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However, we're going to
neglect that effect even

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though if we were to do a
more complicated analysis,

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we would have to change
our model as the object is

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following, so it would
be a multi-stage motion.

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First, at the beginning
with our only velocity

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dependent resistant.

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And then as it gets some initial
speed and it's going faster,

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we change our model.

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That's why the actual problem
can be quite complicated.

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But we're just trying to
keep things simple here.

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And then u of t is square
root of beta mg d of t.

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And then with this change
of variable, my integration,

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remember I have a dv, so I have
to multiply the left side by mg

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over beta.

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And I'm integrating with
a minus sign du times 1

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minus u squared from 0 to
this final value u of t

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And over here I'm just
integrating minus g dt.

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Now again, for
simplicity, I'm going

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to bring this term, the beta
over mg over to the other side.

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So I'll use the magic
of our light board

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by just erasing
that and bringing it

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to the other side, which
makes my life a little easier.

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And now, this interval
can be done by the method

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of integration by parts.

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It's a nice problem in calculus.

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And you can verify for
yourself that the result

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is one half natural
log of 1 plus u over 1

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minus u evaluated at our limits.

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And over here I have minus g
square root of beta over mg.

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Now, once again, for a
little bit of simplicity,

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I'm going to bring the 2
over to the other side.

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And now, I evaluate my limits.

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Now recall that when
you have a minus log,

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we're flipping, because log
of b over a equals minus

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log over ab, so when I
put out my limits in,

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I have natural log of-- now
remember, what are our limits?

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We have 1-- I'm flipping--
minus u is beta mg times v of t.

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And I have the 1
plus beta mg v of t.

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And that's equal to minus as
2g square root of beta mg.

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Now, again, we'll use the
fact that e to log of x is x.

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And so if I
exponentiate both sides,

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I end up with 1 minus the square
root of beta mg d of t over 1

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plus square root
of beta mg v of t

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is equal to exponential
minus 2g beta mg times t.

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And we'll just move that.

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OK, now this is a
little bit of algebra.

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I want to solve for v of t.

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If I bring this
side over to there,

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I'll just do that to make the
first step a little simpler

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to see.

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So we have 1 minus square
root of beta mg v of t

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equals 1 plus square root
of beta over mg v of t times

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e to the minus this factor
2g square root of beta mg t.

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Now this is a lot of
stuff to carry around.

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I'd like to introduce
a constant here,

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tao, which I'm going to find
to be square root of mg beta 1

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over 2g.

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And so this whole term is going
to just be e to the minus tao.

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It's a nice example for you to
work out that the units of tao

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are the units of seconds.

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And that's a little
exercise to work out.

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Now, I just have to
collect my terms.

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And what I'll do is
I'll collect the T

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terms on the right and the terms
that don't have vt on the left.

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So I have 1 minus e to
the minus t over tao

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on the left is equal to 1
plus e to the minus t over tao

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on the right times
beta mg v of t.

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And so I get my
solution, v of t equals

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the square root of mg
over beta times 1 minus

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e to the minus t over tau over 1
plus e to the minus t over tao.

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Well, it's not a
simple solution at all.

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But let's examine when you
have a case like this-- again,

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it would be a nice
exercise to graph this out.

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But right now we're going
to consider the limit

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as t goes to infinity.

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And remember that
e minus t over t

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goes to 0 when t
goes to infinity.

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So we just have 1 over 1.

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And what we get as
t goes to infinity

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is the quantity mg over beta.

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And this is what we call
the terminal velocity.

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Now what does terminal
velocity mean?

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Well, when object is falling
and there's a resistive force,

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as the object falls
faster and faster,

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the resistive force
gets greater and greater

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until if we go back to Newton's
second law and look at it,

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as v gets faster and
faster, eventually

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these two terms are equal.

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And when these two
terms are equal,

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that's the statement that the
right-hand side has to be zero.

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So what we mean by
terminal velocity

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is it's the velocity is no
longer changing in time.

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And then we can
immediately check our work

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by going to Newton's
second law and see what

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that case is if we set this
quantity equal to zero.

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In other words, when
we set mg minus beta v

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squared terminal equal to 0
we can solve for v terminal,

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and we get a square root
of mg divided by beta.

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And that agrees with
our lengthy calculation.

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So we think we're
on the right track.