WEBVTT
00:00:03.940 --> 00:00:08.189
We just arrived this
relation here-- the relation
00:00:08.189 --> 00:00:11.340
between the differential of
the speed of the rocket sled
00:00:11.340 --> 00:00:14.410
and the differential of
the mass of the rocket.
00:00:14.410 --> 00:00:17.830
And we want to ultimately
get the speed of the rocket.
00:00:17.830 --> 00:00:22.950
So we have to apply a technique
called separation of variables
00:00:22.950 --> 00:00:25.464
and then we want to integrate.
00:00:25.464 --> 00:00:27.110
What we're going to
do is, first we're
00:00:27.110 --> 00:00:33.300
going to divide-- multiply
by dt so that falls away,
00:00:33.300 --> 00:00:42.610
and we are left with m
r d v r equals d m r u.
00:00:42.610 --> 00:00:47.100
And we're going to shuffle
the m onto the other side.
00:00:47.100 --> 00:00:49.310
And we're left
with d v r equals--
00:00:49.310 --> 00:00:53.220
u is a constant-- so
that goes up front,
00:00:53.220 --> 00:00:58.510
and then we have d m r over m r.
00:00:58.510 --> 00:01:01.716
That's an equation that
we can integrate now.
00:01:01.716 --> 00:01:05.830
And we can do that.
00:01:05.830 --> 00:01:08.320
And the tricky bit
is that we need
00:01:08.320 --> 00:01:10.930
to take care of the
integration limits here.
00:01:10.930 --> 00:01:16.390
We have v r, and actually
these now are all primes.
00:01:16.390 --> 00:01:20.695
So we have v r going from v0.
00:01:20.695 --> 00:01:24.132
That is our initial
condition here 2vr
00:01:24.132 --> 00:01:34.070
prime of equal vr of t.
00:01:34.070 --> 00:01:39.270
And for the mass, we
have m r prime equals m0.
00:01:39.270 --> 00:01:45.515
Actually, not quite m0, it's
2m0 because the initial mass
00:01:45.515 --> 00:01:48.650
of this is the dry
mass and the fuel mass,
00:01:48.650 --> 00:01:56.400
so that's 2m0, so this
is the initial mass.
00:01:56.400 --> 00:02:04.930
And then we go to m r
prime equals m of vr.
00:02:04.930 --> 00:02:07.400
All right, so let's do that.
00:02:07.400 --> 00:02:16.930
We're going to get vr
minus v0 equals u 1 over m,
00:02:16.930 --> 00:02:21.014
integrated gives us lm, so lm.
00:02:21.014 --> 00:02:28.850
And then we can immediately
do this here over 2m0.
00:02:32.750 --> 00:02:36.810
And well, we
ultimately want this,
00:02:36.810 --> 00:02:38.980
so this is the r
of t, of course.
00:02:38.980 --> 00:02:51.610
And then we got v0 plus
u l n m r over 2 m0
00:02:51.610 --> 00:02:54.090
And that is our equation.
00:02:56.850 --> 00:02:59.530
So what does this
equation tell us?
00:02:59.530 --> 00:03:05.190
M r, the mass of the
rocket is less later
00:03:05.190 --> 00:03:10.600
on than it was before,
which means this term here
00:03:10.600 --> 00:03:14.680
is going to be
less than 0, which
00:03:14.680 --> 00:03:18.130
means the velocity is
our initial velocity
00:03:18.130 --> 00:03:20.900
minus something.
00:03:20.900 --> 00:03:25.480
Which means we have a
decrease in velocity,
00:03:25.480 --> 00:03:29.750
which means my sled will
eventually come to a stop.