Ground Speed and Direction of an Airplane An airplane has an...

Ground Speed and Direction of an Airplane An airplane has an airspeed of 500 kilometers per hour $(\mathrm{km} / \mathrm{h})$ bearing $\mathrm{N} 45^{\circ} \mathrm{E}$. The wind velocity is $60 \mathrm{~km} / \mathrm{h}$ in the direction $\mathrm{N} 30^{\circ} \mathrm{W}$. Find the resultant vector representing the path of the plane relative to the ground. What is the groundspeed of the plane? What is its direction?

Key Concepts

Vector Addition

Vector addition is the process of combining two or more vectors to determine their cumulative effect. In the context of navigation and motion, this involves summing the individual components of different velocity vectors to obtain a resultant velocity vector, which represents the actual path and speed of an object relative to a reference frame, such as the ground.

Resolution into Components

Resolution into components involves breaking down a vector into its horizontal and vertical parts using trigonometric functions. This technique is essential in solving problems involving vectors, as it allows for the independent treatment of motion in different directions, which can be subsequently recombined through vector addition.

Navigational Bearings and Angle Conversion

Navigational bearings are used to specify directions relative to the cardinal points and require careful conversion into standard mathematical angles for calculations. This conversion enables the use of trigonometric functions to extract the appropriate components necessary for vector addition and analysis of motion.

Resultant Velocity and Groundspeed Calculation

The resultant velocity, determined via vector addition, represents the real motion relative to the ground when multiple influences, such as wind and airspeed, are considered. Calculating the magnitude of this resultant vector yields the groundspeed of the object, revealing how fast it is moving over the ground, while its direction provides the bearing of the resultant path.

Transcript

00:01 We have an airplane, i'll call it p, and it is traveling at 500 kilometers per hour and going north, 45 degrees east.

00:12 And so that means, in our picture, we are north and then 45 degrees east, and this is 500, which is also 45 degrees here.

00:24 And that means these components are going to be for a 45 -4590 triangle.

00:30 We end up having this is 500 divided by the square root of two is what the magnitude of each of these is going to be.

00:37 So if i take 500 divided by the square root of two, that comes out to be 355 .5.

00:49 Whoops, 353.

00:51 I'm looking at the calculator wrong.

00:55 353 .555i.

00:57 And it will also be, since we're going over and up, it will be plus 353 .55j.

01:07 Now, the wind is traveling at 60, so we'll say the wind is traveling at 60 kilometers per hour, and it is going north, 30 degrees west.

01:22 So let's just draw a little vector here.

01:25 It's going north and then 30 degrees west so that's coming up like so and i made this way too long in comparison but i just to get my picture in here and i'm going to go head to tail over here in a second and here's a 60 degree angle and we know that this is a 30 degree angle and so this magnitude will end up being 30 kilometers per hour here because opposite in a 30 60 90 triangle when this magnitude is that that is kilometers per hour here with this being 30 or 60 kilometers per hour.

02:05 The short leg is half and then this long leg is 30 times a square root of three and this will actually be a negative component.

02:13 But let's find out what 30 times a square root of three is.

02:18 And that is 51 .96.

02:23 So we can write the components of this factor as the x component.

02:27 Is negative 30 i and then it goes up and that is 51 .96 j so the resultant will be adding these two and so the resultant let's go in red and i'll just call it r as the resultant is i'm going to be 320 .5 .5 i and then when we take that 353 .555 and add the 51 .5 and add the 51 .9 to it, we get the j component is 405 .51, and that is j.

03:06 Now, if we look at this, and if i actually graphed the wind over here, it was supposed to be 30 degrees, and then it would go up a magnitude of 60.

03:19 And so this is our resultant vector, and so we would go over 323 ,000.

03:28 And then we would be going up 405 .51.

03:33 Now, we want to find the speed...

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