WEBVTT
00:00:00.120 --> 00:00:03.520
An airplane is flying on a bearing of one hundred and twenty degrees.
00:00:03.740 --> 00:00:12.560
So we start off looking north as we have here, come round in a clockwise direction of a hundred and twenty degrees; this is the direction that our airplane is flying in.
00:00:14.660 --> 00:00:17.350
It has an airspeed of three hundred miles per hour.
00:00:17.620 --> 00:00:22.760
Now that means that it’s travelling at three hundred miles per hour in relation to the air around it.
00:00:25.290 --> 00:00:30.330
Now the next sentence says that there’s a thirty-mile an hour wind blowing from the north, directly southwards.
00:00:30.330 --> 00:00:35.180
So we’ve got a thirty-mile an hour wind blowing in this direction.
00:00:35.560 --> 00:00:45.380
So if it’s travelling three hundred miles per hour in relation to the wind — relative to the air around it — that air is travelling southwards at thirty miles an hour to start off with.
00:00:45.380 --> 00:00:51.490
So that’s kinda of an additional component of velocity blowing in the southward direction — travelling in that southward direction.
00:00:52.350 --> 00:00:56.270
Now we’ve gotta find the actual bearing of the airplane and its groundspeed.
00:00:58.190 --> 00:01:07.030
So we’re gonna represent all this information in a vector diagram and then we’re gonna use some of our vector knowledge that we’ve gained in order to solve that problem.
00:01:09.460 --> 00:01:11.060
Right, so here’s our diagram.
00:01:11.060 --> 00:01:18.210
We’ve got vector 𝑎 represents the air speed or the air velocity of the aircraft.
00:01:18.580 --> 00:01:21.670
So it’s on a bearing of a hundred and twenty degrees as we said before.
00:01:21.870 --> 00:01:27.970
And the magnitude of vector 𝑎 is going to be three hundred because that’s the speed with which it’s travelling.
00:01:27.970 --> 00:01:31.530
So the magnitude of vector 𝑎 is three hundred moving in that direction.
00:01:32.170 --> 00:01:43.060
Now because the wind is travelling at thirty miles an hour in this direction, the groundspeed is gonna have this extra component of southerly speed to it.
00:01:43.060 --> 00:01:48.750
So the groundspeed is gonna be slightly bigger than that and it’s gonna be heading in a slightly more southerly direction.
00:01:51.080 --> 00:01:58.600
And we’re also trying to find the size of this angle here; call that 𝜃.
00:01:58.950 --> 00:02:07.230
And if we add that 𝜃 to the hundred and twenty degrees we had up here, that would give us the bearing of the groundspeed — the actual bearing — that the-the aircraft is heading on.
00:02:07.900 --> 00:02:11.630
Now what you might see is we’ve got two vectors and we’re trying to find the angle between them.
00:02:11.630 --> 00:02:13.900
So we’re gonna be using vector dot products in here.
00:02:14.280 --> 00:02:17.060
And we’re also gonna be looking at magnitudes of vectors.
00:02:17.060 --> 00:02:19.060
So we’re gonna be using some of those skills as well.
00:02:21.440 --> 00:02:24.580
So let’s try and fill in some of the details about vector 𝑎.
00:02:24.920 --> 00:02:30.430
Well, it’s got an 𝑥-component, which is in this case it’s an easterly direction on our diagram.
00:02:30.430 --> 00:02:34.200
And it’s got a 𝑦-component, which in this case is in a southerly direction.
00:02:34.200 --> 00:02:39.320
So it’s gonna be a negative number; it’s heading downwards in the negative 𝑦-direction.
00:02:40.050 --> 00:02:45.440
We said that the magnitude of our vector was three hundred because it’s travelling at three hundred miles per hour.
00:02:45.470 --> 00:02:51.550
So let’s just do a little bit of work on this triangle that we’ve created before we work out what the 𝑥-component and the 𝑦-component are.
00:02:52.990 --> 00:03:06.870
So I’ve just drawn in my west-, east-, and southern and I filled in this angle here of thirty degrees because the angle between the north and the east direction is ninety degrees, which leaves thirty degrees for this angle here.
00:03:08.860 --> 00:03:12.180
So just looking at the cosine of that then, we got a right-angled triangle.
00:03:12.180 --> 00:03:18.600
The cosine of thirty degrees is the adjacent side over the hypotenuse; so that’s 𝑎𝑥 over three hundred.
00:03:18.850 --> 00:03:24.340
So we can rearrange that to work out what 𝑎𝑥 is, simply by multiplying both sides by three hundred.
00:03:25.590 --> 00:03:28.980
So that’s 𝑎𝑥 is three hundred cos thirty.
00:03:29.230 --> 00:03:32.050
And then cos thirty of course is root three over two.
00:03:33.820 --> 00:03:41.940
So when we work all that out, the-the easterly component of 𝑎’s velocity is a hundred and fifty root three miles per hour.
00:03:42.740 --> 00:03:45.380
Okay, let’s look at 𝑎𝑦 now — the 𝑦-component.
00:03:46.890 --> 00:04:00.380
And sine of thirty is the opposite side of the hypotenuse which is 𝑎𝑦 over three hundred, which again multiply both sides by three hundred and we get three hundred sin thirty.
00:04:00.960 --> 00:04:04.580
Now sin thirty is a half, so that equals a hundred and fifty.
00:04:06.260 --> 00:04:17.550
So the airplane’s speed if you like in the easterly direction is a hundred and fifty root three and in the southerly direction is a hundred and fifty miles per hour.
00:04:18.190 --> 00:04:21.140
So we’ve worked out what the components of the 𝑎 vector are.
00:04:22.510 --> 00:04:29.760
Now the one thing that we do have to watch out for is that remember the-the positive 𝑦-direction would be going in a more northerly direction.
00:04:29.760 --> 00:04:33.520
But because we’re heading towards the south, we’ve got a negative 𝑦-component.
00:04:33.520 --> 00:04:37.040
So that a hundred and fifty miles per hour is in the southerly direction.
00:04:37.040 --> 00:04:44.850
So on our scale, the 𝑎 vector is gonna be this here: a hundred and fifty root three, negative one hundred and fifty.
00:04:48.380 --> 00:04:54.820
So moving on to the groundspeed vector, we can see that the horizontal component is exactly the same as the airspeed.
00:04:55.160 --> 00:05:00.660
The extra wind was only blowing in the southerly direction; so it’s not-not affecting the horizontal component.
00:05:00.660 --> 00:05:06.480
So the the horizontal component — the 𝑥-component — of the groundspeed is the same as for the airspeed.
00:05:07.040 --> 00:05:13.720
Now with the southerly component, the 𝑦-component, we’ve got this extra thirty miles an hour kicking in.
00:05:13.720 --> 00:05:20.180
So we’ve got whatever the airspeed was up here, but we’re adding an extra thirty miles an hour to that.
00:05:20.180 --> 00:05:30.280
So the-the-the groundspeed 𝑦-component is gonna be a hundred and fifty same as airspeed plus the extra thirty is a hundred and eighty miles per hour.
00:05:32.140 --> 00:05:52.730
And again because that is in a southerly direction, not a northerly direction, it’s the negative 𝑦-component, so it’s gonna be minus a hundred and eighty miles an hour so that our 𝑔 vector is a hundred and fifty root three for the 𝑥-component same as the airspeed and negative a hundred and eighty that extra thirty miles an hour in the 𝑦-component.
00:05:55.330 --> 00:06:01.140
So we’ve worked out our velocity vectors then for the airspeed and groundspeed 𝑎 and 𝑔 vectors.
00:06:01.510 --> 00:06:06.970
And what we now need to do is work out what that- the magnitude of that groundspeed is.
00:06:07.000 --> 00:06:10.550
So we’re going to try and find the magnitude of 𝑔.
00:06:11.960 --> 00:06:17.370
And to do that remember we square the 𝑥-component and we add that to the square of the 𝑦-component.
00:06:18.360 --> 00:06:35.650
And calculating those, we get the square root of ninety-nine thousand nine hundred, which is equal to thirty root one one one if you wanna be exact about it or to one decimal place three one six point one miles per hour.
00:06:38.290 --> 00:06:44.810
So the magnitude of the groundspeed is three hundred and sixteen point one miles per hour to one decimal place.
00:06:48.140 --> 00:06:49.820
So that’s our groundspeed then.
00:06:49.850 --> 00:06:56.560
Now what we need to do is work out the size of this angle here so that we can add it to a hundred and twenty degrees and get the size of our bearing.
00:06:58.850 --> 00:07:06.380
And to work out the angle, we’re going to use the dot product of the unit vectors in the direction of the airspeed and the groundspeed.
00:07:06.380 --> 00:07:11.750
So 𝑎 over the magnitude of 𝑎 dot 𝑔 over the magnitude of 𝑔 in this case.
00:07:13.940 --> 00:07:19.640
So the magnitude of 𝑎 is the components squared and added together and take the square root of that.
00:07:19.640 --> 00:07:23.490
We’ve already worked out the magnitude of 𝑔; so we can now just simplify that down.
00:07:24.930 --> 00:07:33.490
And to work out these dot products, I’m gonna take this component and multiply it by that component and add it to this component multiplied by this component.
00:07:36.290 --> 00:07:38.080
So that’s what that gives us.
00:07:38.280 --> 00:07:46.740
Remember because we’re doing dot products, we mustn’t use the-the cross-multiplication sign; we need to be consistent and use those dots for multiplying the numbers together.
00:07:49.330 --> 00:08:00.090
So just got a common denominator that already multiplying those numbers out and adding them together gives us twenty-one over two root one one one.
00:08:00.370 --> 00:08:07.480
So what we’re saying is that this cos 𝜃 here is equal to twenty-one over two root one one one.
00:08:07.480 --> 00:08:11.350
So if I do cos to the minus one of this, it will tell me what 𝜃 is.
00:08:13.030 --> 00:08:17.180
And to one decimal place that gives us four point seven degrees.
00:08:17.530 --> 00:08:30.390
Now it’s just worth mentioning at this point I haven’t been rounding these numbers as I’ve been going along; I’ve been trying to keep them in a surd format — root format — just to keep the accuracy in the question as long as possible.
00:08:31.190 --> 00:08:38.210
So it’s not always possible to do that because you know some numbers don’t come out quite as easily; the calculator can’t handle them in that format.
00:08:38.210 --> 00:08:41.600
But if they can, it’s best to work in that format for as long as possible.
00:08:43.550 --> 00:08:46.420
Now remember in our question, so we’ve just worked out this bit here.
00:08:46.690 --> 00:08:50.720
So we got to add that to a hundred and twenty to work out what the actual bearing was.
00:08:51.820 --> 00:09:00.930
So that’s gonna give us an answer of a hundred and twenty-four point seven degrees So there we have it.
00:09:00.980 --> 00:09:12.580
Our answers were three hundred and sixteen point one miles per hour to one decimal place for the groundspeed and a hundred and twenty-four point seven degrees to one decimal place for the bearing.
00:09:14.750 --> 00:09:17.550
So let’s just kinda recap some top tips as we go.
00:09:17.760 --> 00:09:22.460
Always worth doing a diagram, so that definitely helped us doing a diagram.
00:09:22.750 --> 00:09:28.310
We represented the airspeed and the groundspeed as vectors, so that turned out to be quite useful.
00:09:28.600 --> 00:09:36.250
We were able to work out the magnitude of vector 𝑔 fairly easily just using effectively the Pythagorean theorem.
00:09:36.660 --> 00:09:46.480
We were trying to keep our values in surd format — in this root format — as far as possible to keep the numbers as accurate as possible as well.
00:09:46.770 --> 00:10:00.310
And we were also able to use the-the cos 𝜃 equals the dot product of the unit vector in each direction to work out the angle between those two vectors.