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Vectors question

  1. Feb 20, 2016 #1
    • Moved from a technical forum, so homework template missing
    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.

    ________________________________________________________________________________________

    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. Then using cosine and sine laws you can find the magnitude and angle of the resultant vector.

    ________________________________________________________________________________________

    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).

    ________________________________________________________________________________________

    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).
     
  2. jcsd
  3. Feb 20, 2016 #2
    If the bearing is x degrees, does this mean that it is x degrees measured clockwise relative to North?
     
  4. Feb 20, 2016 #3

    A.T.

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    You think something is correct, because you found it online?
     
  5. Feb 20, 2016 #4
    @Chester, if it's "true bearing" then it's from North. In this case it was "true bearing," so that's why it's from North.

    @A.T, I wouldn't think it's correct if I found it online from 1 source. I checked multiple sources and found the same answer, and that's how I concluded it probably is the right answer.
     
  6. Feb 20, 2016 #5

    Ray Vickson

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    You still left one part of Chestermiller's question unanswered: do you mean 120 degrees clockwise or counterclockwise (from North)?
     
  7. Feb 20, 2016 #6
    Yes, it's clockwise.
     
  8. Feb 20, 2016 #7

    phinds

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    Please draw a vector diagram to show exactly what you are talking about.
     
  9. Feb 20, 2016 #8
  10. Feb 20, 2016 #9
    OK. This is the picture I envisioned. Now, do you know how to resolve these vectors into their components in the x and the y directions? If so, what are the x and y components of each of the vectors?
     
  11. Feb 20, 2016 #10
    Hi, I actually don't know how to do that since I never used that method where I divided the vectors into their x and y components. As you can see in the picture, there's one triangle, so at this point what I would do is simply find the magnitude of the resultant by using the cosine law. That would be as follows-

    r^2=60^2+550^2-2(60)(550)*cos40
    r^2=306100-66000*cos40
    r^2=306100-50558.93
    r^2=255541.07
    r=505.51

    Therefore, the magnitude of the resultant would be 505.51 km/h. This, of course, is according to this method. Using the other method (which I showed in my initial post), the magnitude of the resultant came out to be 597.2 km/h. That's what my confusion was. The two methods give me different answers. When I checked online, some sources showed that the answer was 505.51 km/h (as found by method 1) and some sources showed that the answer was 597.2 km/h (as found by method 2). So, apparently both answers are right.
     
  12. Feb 20, 2016 #11

    phinds

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    And do you seriously believe both answers CAN be right?
     
  13. Feb 20, 2016 #12
    I get an angle between the two vectors of 140 degrees if I place the vectors tail-to-tail.

    If I take the components of the two vectors in the x and y directions, I get:

    Plane:
    x component = 550 cos 10
    y component = 550 sin 10

    Wind:
    x component = - 60 cos 30
    y component = 60 sin 30

    What do you get for the resultant x component?
    What do you get for the resultant y component?
     
  14. Feb 20, 2016 #13
    No both can't. That is the confusion. When I was checking for the answer online, I was finding both answers (550.51 km/h and 597.2 km/h), and that really threw me off.
     
  15. Feb 20, 2016 #14
    The 597.2 gives the magnitude of the difference between the two vectors. This is what I get when I use the 140 degrees, laying the vectors tail to tail. The 505.51 (which is the correct answer) gives the sum of the two vectors. This is what you got when you used the 40 degrees, which is consistent with laying the vectors tail to head to add them. Doing the problem in terms of the components gives the resultant of the two vectors as 505.51 also. So the 597.2 is the incorrect answer.
     
  16. Feb 20, 2016 #15
    I actually just asked someone about this and they said that there was an error made in the following solution:

    ____________________________________________________________________________________________________________________________________

    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
    ____________________________________________________________________________________________________________________________________

    The error was the "120" degrees in the initial statement "=[550(cos80+i*sin80)] + [60(cos120+i*sin120)]." That 120 degrees should have been 300 degrees. The reason being, due to the context of the question, wind is considered a back bearing since it says in the question "the wind is blowing at 60 km/h FROM a bearing of 120 degrees." Due to the term "from" used in the question, that means wind is considered a back bearing. For that reason the back bearing formula would be applied in this case: back bearing=(180 degrees+bearing)=(180+120)=300. Therefore, the 120 in the formula should be replaced with 300. When I do that the answer works out to be 505.50 km/h. The proper solution can be seen below:

    ____________________________________________________________________________________________________________________________________

    Vr=Vp+Vw
    =[550(cos80+i*sin80)] + [60(cos300+i*sin300)]
    =[95.50+541.64i] + [30-51.96i]
    =125.50+489.68i

    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 (125.50^2+489.68^2)
    =505.51

    TanTheta=489.68/125.50
    Theta=75.63

    ____________________________________________________________________________________________________________________________________
     
  17. Feb 20, 2016 #16
    This is what I get doing the problem in component form.

    Chet
     
  18. Feb 20, 2016 #17
    Can you explain a bit about the whole "back bearing" concept I mentioned. I'm a bit confused about when to use back bearing. In the case of the plane vector in the question, the forward bearing was used, and in the case of wind vector in the question, the back bearing was used.
     
  19. Feb 20, 2016 #18
    The key words are "blowing from." That means that the wind is blowing from the 120 degree direction. You have correctly shown the wind direction in your diagram.
     
  20. Feb 20, 2016 #19
    So whenever it says "blowing from" I use the back bearing for that formula (method 2), which in this case was (180+120)=300. And, if, for instance it said the wind is "blowing at" a bearing of 120 degrees then I would use the 120 rather than the 300?
     
  21. Feb 20, 2016 #20
    Yes. But ask yourself this: how important is it to remember this? How many times are you going to run into wind and plane problems in real life?
     
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