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We just arrived this
relation here-- the relation

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between the differential of
the speed of the rocket sled

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and the differential of
the mass of the rocket.

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And we want to ultimately
get the speed of the rocket.

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So we have to apply a technique
called separation of variables

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and then we want to integrate.

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What we're going to
do is, first we're

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going to divide-- multiply
by dt so that falls away,

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and we are left with m
r d v r equals d m r u.

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And we're going to shuffle
the m onto the other side.

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And we're left
with d v r equals--

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u is a constant-- so
that goes up front,

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and then we have d m r over m r.

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That's an equation that
we can integrate now.

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And we can do that.

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And the tricky bit
is that we need

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to take care of the
integration limits here.

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We have v r, and actually
these now are all primes.

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So we have v r going from v0.

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That is our initial
condition here 2vr

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prime of equal vr of t.

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And for the mass, we
have m r prime equals m0.

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Actually, not quite m0, it's
2m0 because the initial mass

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of this is the dry
mass and the fuel mass,

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so that's 2m0, so this
is the initial mass.

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And then we go to m r
prime equals m of vr.

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All right, so let's do that.

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We're going to get vr
minus v0 equals u 1 over m,

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integrated gives us lm, so lm.

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And then we can immediately
do this here over 2m0.

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And well, we
ultimately want this,

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so this is the r
of t, of course.

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And then we got v0 plus
u l n m r over 2 m0

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And that is our equation.

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So what does this
equation tell us?

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M r, the mass of the
rocket is less later

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on than it was before,
which means this term here

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is going to be
less than 0, which

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means the velocity is
our initial velocity

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minus something.

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Which means we have a
decrease in velocity,

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which means my sled will
eventually come to a stop.