Vectors world problem

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Math4ever

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Hello guys,


I have a vectors word problem and I found 2 different ways to solve the same problem but I'm getting different answers. Apparently, both answers are correct since I've looked for the answer online and I found both answers from different sources, so I'm really confused now.


Question: An airplane is flying at 550 km/h on a heading of 080 degrees. The wind is blowing at 60 km/h from a bearing of 120 degrees. Find the ground velocity (resultant vector) of the airplane.



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Method 1: First you would draw the vector diagram. You would draw a y-axis and a x-axis. Then from the origin you would draw the plane vector at an angle (from north) of 80 degrees and this vector would be 550 km/h in magnitude. Then from the origin you would draw the wind vector at an angle (from north) of 120 degrees and this vector would be 60 km/h in magnitude. Then you would connect the outside ends (away from the origin) of each vector together. This side would be the resultant. This will form a triangle. Using cosine law you will find the resultant magnitude and then using the sine law you will find the resultant angle.


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Method 2: I found this method online and I don't even quite understand it myself but apparently it works. All values used in this method are from the question.
Vr=Vp+Vw
=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]
=[95.50+541.64i] + [-30+51.96i]
=65.50+593.60i


Since this is in a+bi form you will now convert it to polar form. You will do this using the Pythagorean theorem and tan.


r=square root (65.50^2+593.60^2)
=597.2


TanTheta=593.60/65.50
Theta=83.70


Therefore, the magnitude of the resultant would be 597.2 km/h and the angle would be 83.70 degrees (from north).


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Anyways, these are the two methods. If you solve this using method 1 you will get a different answer than method 2. However, both answers are apparently correct since I checked for the answer to this question online and I found both answers. I'm confused now about which answer would be the correct one (using method 1 or method 2).



 
Your method 2 answer is obviously wrong. The wind from 120 has a headwind component - it will result in a ground speed which is lower than the airspeed. What was your method 1 answer? What is the angle between the airspeed vector and the wind vector?
 
Last edited:
Note that the wind is blowing FROM 120 degrees....which adds speed to the aircraft....

zzzzzzzzzzz.jpg
 
Math4ever said:
An airplane is flying at 550 km/h on a heading of 080 degrees. The wind is blowing at 60 km/h from a bearing of 120 degrees.
Click to expand...
What does "bearing" mean here? Where are the angles measured from?
Also, what answer do you get using the 1st method?
Do you agree with the diagram posted by @Pavlov ?
 
blamocur said:
What does "bearing" mean here? Where are the angles measured from?
Also, what answer do you get using the 1st method?
Do you agree with the diagram posted by @Pavlov ?
Click to expand...
Note that @Pavlov has resurrected an 8-year-old question. The OP won't respond.
Pavlov said:
Note that the wind is blowing FROM 120 degrees....which adds speed to the aircraft....

View attachment 38212
Click to expand...
Your picture is approximately correct, once I realize your N is on the right and angles are measured counterclockwise; but it doesn't match your calculation, in which you used +60 instead of -60. Notice that the speed is not increased by the wind, which is in fact more of a headwind.
Math4ever said:
I have a vectors word problem and I found 2 different ways to solve the same problem but I'm getting different answers. Apparently, both answers are correct since I've looked for the answer online and I found both answers from different sources, so I'm really confused now.
Click to expand...
And of course, nothing on the internet is wrong ...

Math4ever said:
Question: An airplane is flying at 550 km/h on a heading of 080 degrees. The wind is blowing at 60 km/h from a bearing of 120 degrees. Find the ground velocity (resultant vector) of the airplane.
Click to expand...
Normally the heading and bearing mean the angle clockwise from north.

The direction opposite a bearing of 120 is -60, not +60.

Here is a more accurate picture (B is bearing, W is wind, north is up), if I didn't make a mistake of my own:

1719000693137.png

Since both "method 2" and @Pavlov are using a coordinate system where N is +x and E is presumably +y, the work (when done with the right angles) in both would give the correct answer, though perhaps misleading on a map. And @Pavlov's correction about the word "from" is right.
 
Dr.Peterson said:
Note that @Pavlov has resurrected an 8-year-old question. The OP won't respond.

Your picture is approximately correct, once I realize your N is on the right and angles are measured counterclockwise; but it doesn't match your calculation, in which you used +60 instead of -60. Notice that the speed is not increased by the wind, which is in fact more of a headwind.

And of course, nothing on the internet is wrong ...


Normally the heading and bearing mean the angle clockwise from north.

The direction opposite a bearing of 120 is -60, not +60.

Here is a more accurate picture (B is bearing, W is wind, north is up), if I didn't make a mistake of my own:

View attachment 38216

Since both "method 2" and @Pavlov are using a coordinate system where N is +x and E is presumably +y, the work (when done with the right angles) in both would give the correct answer, though perhaps misleading on a map. And @Pavlov's correction about the word "from" is right.
Click to expand...
Yah.......It was my first post on this site and I found the question looking for something else and did not realize it was 8 years old !!! Ooops ! (and I made a math errror and ADDED the vertical component of wind speed instead of subtracting it ...D'oh! ) .... Thanx for the correction..... I posted a corrected version below..... ~ P
 
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