00:00Hi guys, so here we're
00:03going to look at a
00:03nice application of vectors and
00:05that is in kinematics. So
00:09let me just introduce you
00:11to this using this beautiful
00:13picture of an airplane that
00:14I've drawn. We can model
00:19the position of an airplane
00:24or a boat or anything
00:25moving in straight lines using
00:28vector equations. So for example,
00:33this plane has position vector
00:38750 plus t times minus
00:40543, where t is the
00:43time, now that could be
00:44in hours or seconds, I
00:47think in this particular case,
00:49we're going to assume it's
00:51seconds. Now, what's happening is
00:54the plane
00:56the plane has taken off.
00:59So let's say this is
01:00the ground here. So his
01:04position vector is 7, 5,
01:070. So this point is
01:09like 7, 5, 0. Note
01:110 because that is, well,
01:15in this particular example, the
01:16z axis is our kind
01:20of vertical axis in which
01:22the plane
01:24takes off and 7 .5.
01:28This is like your x
01:30-coordinate and y -coordinate on
01:31the ground, but zero or
01:33the z -coordinate is how
01:36far up in the air
01:37he is. So after one
01:39second he is three meters
01:41in the air. After 10
01:43seconds he's 30 meters in
01:44the air and these two
01:46describe ways and the x
01:48-y plane. So what happens
01:51is every
01:52Three second he travels this
01:56much negative five in the
01:58x direction, four in the
02:00y direction, and three in
02:01the z direction. So he's
02:03traveling this much, this vector
02:06per second. So every second
02:08we're going to add this
02:09on. So think about what
02:11does that mean when you
02:12travel at a certain amount
02:14or when your displacement is
02:16something per second? It means
02:19it's this is your
02:20velocity. So the direction vector,
02:22when we write a vector
02:25equation of a line like
02:28this where the parameter is
02:30actually the time, this is
02:32the velocity. And then the
02:35magnitude of this vector, the
02:38magnitude of the direction vector,
02:44which is, it gives us
02:47the speed.
02:48The speed, remember the speed
02:50is a scalar, so we
02:52don't need the direction, so
02:53the speed is just five
02:55squared plus four squared plus
02:57three squared, which is square
03:01root of 25 plus 16
03:04is 41 plus nine, which
03:09is 50, exactly, square root
03:11of 50. So he's going
03:14at a speed of,
03:16square root of 50 meters
03:18per second. So I don't
03:20know if that's fast for
03:21an airplane or not. Let
03:22me think that would be
03:23approximately seven meters per second.
03:27That's probably not that fast
03:29for an airplane, but whatever.
03:31It's just an example. So
03:32that's what it means. That's
03:33kind of his starting point.
03:35That's his velocity. And then
03:38the whole equation gives us
03:40his position after any amount
03:42of time. So if I
03:43said to you, well, what's
03:44the plane
03:44position after, I don't know,
03:49a hundred seconds, you'd sub
03:50in a hundred there and
03:52you get his position vector.
03:55This obviously will model the
03:57plane from when he takes
03:58off until he reaches his
04:00kind of maximum altitude or
04:03whatever and then obviously it
04:05changes this. The Z component
04:12would be zero again because
04:14he's just gonna maintain that
04:16altitude. Okay, that's a plane
04:20in 3D. This example here,
04:22I'm actually gonna look at
04:23boats, which is more in
04:24two dimensions. So it says,
04:27ship A leaves a port
04:28located at the origin and
04:30moves with a velocity 3i
04:32plus 5j. Guys, whenever I
04:34have a question, whenever I
04:38have a question, I like
04:40to whatever question. I like
04:43to draw a diagram just
04:45to see what's actually happening.
04:47Let me get rid of
04:48this. I want to do
04:51this in black. Imagine I
04:59have... here's my... whatever it
05:02is my plane, my set
05:05of axes, and this ship
05:08is leaving the
05:08and he's moving with a
05:10velocity of three i plus
05:11five j. So let's say
05:12that's three i plus five
05:17j, something like this. He's
05:19moving like this. Fine. That
05:23direction. And then ship b
05:25starts 18 kilometers north. So
05:29he is moving while he's
05:33starting up here somewhere. 18
05:35kilometers north. So directly
05:36north and he's moving it
05:394i minus 2j so let's
05:41go 4 minus 2, 4
05:48minus 2 something like that.
05:53Okay so he's going that
05:54direction, he's going that direction.
05:56It says find the distance
05:57between the ships after two
06:00hours. Okay so
06:04Shabay leaves the port and
06:06he's moving a velocity of
06:083i plus 5j. So ship
06:10A, we can model ship
06:12A's path using this vector
06:15equation, or equals position vector
06:180, 0, he started at
06:200, 0 plus t times
06:24because we're looking at its
06:31kilometers per hour.
06:33Every hour he moves, every
06:36hour he is going to
06:38move 3i plus 5j, so
06:42this becomes 3 3 5.
06:46So that's ship A's position
06:50vector after a given time.
06:53And ship B's position after
06:56a given time will be
06:58or equal. So he starts
07:00at z
07:01zero 18 because he's zero
07:05along the x -axis and
07:0718 up here plus t
07:10times and his velocity is
07:11four minus two, four minus
07:14two like this. So the
07:18question says find the distance
07:19between the ships after two
07:21hours. So after two hours,
07:25after two hours, ship A.
07:29A, what is ship A's
07:33position vector? Well, ship A's
07:35position vector is 0, 0
07:38plus 2 times because it's
07:41now 2 hours, so T
07:42is 2. 2 times 3,
07:445, which is just 6,
07:4910. So his position vector
07:50is 6, 10. And ship
07:53B, ship
07:57B's position vector is going
07:59to be 0 .18 plus
08:032 times 4 negative 2,
08:07which gives us 8, 8,
08:10and then 18 minus 4
08:14is 14, 8, 14. That's
08:18his position vector. That's his
08:20position vector. Let me draw
08:21another little graph.
08:25just to show you what's
08:26happening here. So he is
08:28at this guy's at 6
08:3110. So let's say 6
08:3410. Something like this. That's
08:38his point is 6 10
08:41and ship B. His point
08:44is 8, 8, 14. So
08:48he's here at 8, 14.
08:50So now we just have
08:51those are the points we
08:52now
08:53just have to find the
08:54distance between two points, which
08:58is, we know this formula.
09:00Distance is equal to the
09:04square root of, and it's
09:06this minus the squared plus
09:08this minus the squared. So
09:09it's eight minus six squared
09:11plus 14 minus 10 squared.
09:16This is in the formula.
09:17But look, guys, if you
09:18don't know what I'm doing,
09:20but hopefully, you know, the
09:21distance
09:21since between two points formula
09:22by now. So this is
09:23going to be four, let's
09:29two squared plus four plus
09:33four squared, which is 16,
09:38giving us root 20 kilometers
09:42exactly. Okay, that's it, that's
09:47my lesson on vectors with
09:49kinematics. Obviously the main things
09:52to note is we can
09:54use the vector of equation
09:55of a line to model
09:57the path of some object,
10:03be it a plane or
10:04a ship or whatever it
10:05might be. We can do
10:06it in three dimensions and
10:08two dimensions. The position vector
10:10is kind of its starting
10:12point because it's like when
10:14t is zero. The initial
10:16is initial position.
10:17And then because I'm adding
10:20the direction vector, or I'm
10:22adding a direction vector for
10:23every given time, whether it's
10:25a second or a kilometer,
10:27that is by definition the
10:29velocity and the magnitude of
10:31the velocity gives us the
10:33speed. That's it. Hope that
10:36all made sense and I'll
10:38see you in the next
10:38lesson.