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A jet airliner moving initially at 538 mph (with respect to theground) to the east moves into a region where the wind is blowingat 937 mph in a direction 23â—¦ ...

Question

A jet airliner moving initially at 538 mph (with respect to theground) to the east moves into a region where the wind is blowingat 937 mph in a direction 23â—¦ north of east. What is the new speedof the aircraft with respect to the ground? Answer in units ofmph.

A jet airliner moving initially at 538 mph (with respect to the ground) to the east moves into a region where the wind is blowing at 937 mph in a direction 23â—¦ north of east. What is the new speed of the aircraft with respect to the ground? Answer in units of mph.



Answers

A jet airliner, moving initially at 300 mi/h to the east, suddenly enters a region where the wind is blowing at 100 mi/h toward the direction 30.0° north of east. What are the new speed and direction of the aircraft relative to the ground?

In this video, we will determine the magnitude and direction of a plane's velocity relative to the ground. If it is flying with an air speed of 230 MPH, with a compass heading of 2 85 degrees and there is a wind of 35 MPH In the direction of 260°. The first thing we can do is draw these two victors and they are drawn here with compass is the degrees are measured from due north, So this is 285 And this is to 60. So the vector you here is the plans air speed and the vector v is the wind. And this due west is 270°. So the first thing we can do to solve this problem is find the values of these two angles here and here, which we can call Alpha and Beta. Because This line here is 270° Alpha is equal to 2 85 minus 2 70 Which is 15° And beta is equal to 2 70 -260, which is 10°. Next we can find the x and Y components of each vector, which we can use find by using the equation. The X component is equal to the magnitude times co. Sign it up and the Y component is equal to the magnitude times signed. So in this case the magnitude or the X component of the vector U. Is equal to 2 30 times co sign of 15 degrees Which is to 22 And the y component is to 30 terms sign of 15° which is 60. We can do the same for the x. and y components of the wind with the magnitude of 35 and the angle better and get that the X component is 34. In the wild component is six and we can put a negative six there because it is in the opposite direction of the Y component of the airspeed. Next we can add these components together to find the X and Y components of the final result in factor So to 22 plus 34 Means the x component is 256 and 16 -6 is the excellent. Are the white 154. So the final vector is 256 comma 54 with 2 56 being The length in this direction and 54 being length in that direction. And so then to find the magnitude of this vector, we can use the pythagorean theorem where the magnitude is equal to the square root of the X component squared plus the y component squared which is to 61.63 mph here and to find the angle or the direction of the spectrum, we can use the formula the inverse tangent of the Y component over the X component is equal to the angle, And we get that the angle is 11.91°. However, that angle is relative to this line here, because the resultant vector will look someone like that and so to find the direction relative to north, We can add 270° to that And get the angle to any 1.91°. And these are final answers.

Okay, we have an airplane that heads northeast at an air speed of 700 km/h. There's a wind blowing from the east. So when it says from the east, make sure that you're going or from the west make sure you're going towards the east and that's at 60 kilometers per hour. And we are going to be finding the direction and the relative speed of that plane. So first of all, we're going to write out our component form of our velocity vectors. So the plane will be written as 700 co sign of 45 in the eye and also 700 of sign of 45 in the J. And we all know that co sign and sign of 45 or both. That squirt 2/2. Okay, now the wind is just going to be in the positive extractions. We have a 60 I but plus a zero J. So take that 700 multiply it by square root of 2/2. We get that 350 times the square root of two. We add that to our 60. We have our eye component and then we just have the 350 square to for the J. component to find our magnitude. We're going to take each of those components and we are going to square them some them and then take the square root That gives us a velocity. Um that has a magnitude of 743.63, km/h. Now we're also going to find the direction so we want to be able to find the angle here. So we're gonna do the tan negative one of our velocity of ry component over the X. Component. We find that we get an angle of 41.8-9. So considering that you would add your vectors tip to tail, we would consider that our final vector should Be less than 45 and longer. So our answers do make sense.

So the velocity of the air is in north east direction. So the horizontal component is three a over route to and the vertical company It is v A over to So doesn't velocity of the plane with restaurant the ground? It was velocity of the played in still air plus velocity off the ear and is a vector some. So from here we could find that with us be of the they just make the ground with them out to be, since the velocity of the plane is in the north. So that's that without skipped a plane. So they speak arms V A over you do square does V a over route to cast a VP squared my road. He was 65 over route do 65 over route to the US Phoebe east to 35 squared. Give me that by our and that comes out to be 1 18 Uh, that to be 1 94 Why? In 53 you help me. Our

Okay, we have a plane that's flying at 340 mph at a heading of 124°. So we'll just kind of write that out in terms of our vectors. And then we have a wind of 45 mph blowing from the West. Now be careful with this from the West means towards the east and so that is going to be at an angle of zero because it's just gonna be straight, so it's 45 MPH. Okay. So what we're gonna do is we're going to come up with are two separate vectors. And then to find the resultant of the two will be adding the vectors together but has to be in their component form. So we'll have our 340 times are co sign of 124 And then we'll have our 340 times the sine of 124. Now when we place them in our calculator really expect that first guy to be negative because we're in quadrant two. But the second one to still be positive. Now the wind is going straight at a 40 a zero. So it's going to be 45 in the eye and zero in the J. If you use co sign of zero you're just going to get a value of one and sign of zero you will get a value of zero. So you can still use your co sign and sign but it's obvious what our result is going to be there. Okay now I'm gonna add my columns so I'll add my 340 co sign over 1 24 2 45. Now as I said before that was a negative number and then plus a positive that's going to be a little bit smaller negative number. And then um we can also put our other value in now for um to help you out here. Um These numbers when you put them into your calculator, um they go a lot farther than one decimal place and we shouldn't be rounding until the very end. So also cuts down on work is if as soon as you find your eye component, if you were to store it in A and as soon as you find your um why components stored and be then you can use those and remember um on the calculator is your store button is right above your on and then you can do like alpha and alpha B. To get those values in. So now when we find our magnitude we can use the A. And the B. And you're not having to re type all of that and make sure you have apprentices and everything for your negatives. So this ends up being 317 mph. And then I look at my vector and my vector is smaller than the 340 that I started with. And so I feel good about my answer.


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