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Math Help - vectors: HELP!!!!!

  1. #1
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    Question vectors: HELP!!!!!

    I am soooo CONFUSED!

    1. An airplane is flying in the direction South 32 degrees East (S32E) with an airspeed of 860 kilometers per hour. Because of the wind, its groundspeed and direction are respectively 800 kilometers per hour and South 40 degrees East (S40E). Find the direction and speed of the wind.

    2. An airplane's velocity with respect to the air is 580 miles per hour, and its heading is North 60 degrees West (N60W). The wind, at the altitude of the plane, is from the southwest and has a velocity of 60 miles per hour with a bearing of North 45 degrees East (N45E). Draw a diagram to represent the problem. What is the true direction of the plane, and what is its speed with respect to the ground?

    HELP!!!!!
    Last edited by iheartthemusic29; February 19th 2008 at 04:31 PM. Reason: new information
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  2. #2
    Newbie DavePercy's Avatar
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    An airplane's airspeed vector plus the wind's velocity vector gives the plane's groundspeed vector. To add vectors, break them down into parts that line up with the coordinate system, add the horizontal components and the vertical components, then convert back to a speed and direction.

    For the first question, use variables for the wind's speed and direction and set up equations with the components of the plane's airspeed plus the components of the wind speed equal to the components of the ground speed, then solve for the wind speed.

    The second question is the same, but easier since you're just adding the airspeed to the windspeed (adding as vectors, not just adding the magnitudes).
    I'm not sure how much detail your teacher wants for the diagram, but either draw both vectors in standard position (tail at the origin) and use the parallelogram method to draw the resultant, or draw one vector in standard position with the other vector's tail at the first vector's head, and draw the resultant from the tail of the first to the head of the last.


    If this isn't enough information or I worded something confusingly then let me know and I can go through the formulas.
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  3. #3
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    Question Well...

    So, can I use the distance formula? Or am I looking at this the wrong way? I'm still a little puzzled.
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  4. #4
    Newbie DavePercy's Avatar
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    The distance formula is part of it, but you need to use a little trig first. There are two formulas for turning a vector's magnitude and angle into components:

    V_x = V cos\; \theta

    V_y = V sin\; \theta

    EDIT: I had used the wrong angles before, I've fixed this.

    So in question 2 (I'm assuming that N60W means the plane makes a 60 degree angle with the north, axis, correct me if I'm reading this wrong. If this is the case, the plane's angle with the east axis is 90+60=150.)

     V_x = 580 \;cos\; 150 = -502.3

     V_y = 580 \;sin\; 150 = 290

    This means that traveling at 580kph at an angle of 120 degrees with the east axis is the same as traveling 290kph west and 502.3kph north--relative to the air. The air is also moving relative to the ground, from the southwest, which is the same as to the northeast, at 60kph.

     V_x = 60 \;cos\; 45 = 84.9

     V_y = 60 \;sin\; 45 = 84.9

    So, the air is moving 84.9kph east and 84.9kph north relative to the ground. If we add Vx of the plane relative to the air and Vx of the air relative to the ground we get Vx of the plane relative to the ground; the same is also true for Vy.

     V_x = -502.3 + 84.9 = -417.4

     V_y = 290 + 84.9 = 374.9

    Now we know that the plane is moving 417.4kph west relative and 374.9kph north relative to the ground; we want to convert this back to a speed and angle:

     V = \sqrt{{V_x}^2 + {V_y}^2} = \sqrt{(417.4)^2+(374.9)^2} = 561

    \theta = tan^{-1}{\frac {V_y} {V_x}} = tan^{-1}{\frac {374.9} {417.4}} = 41.9

    (Note that tan^{-1} is the inverse tan, also sometimes called arctan; it's not a reciprocal, its an inverse function.)

    So the plane's speed relative to the ground is 561, and the angle it travels relative to the ground is northwest, 41.9 degrees with the east axis. N48.1W (This is all assuming that I correctly understood what you meant by "North 60 degrees West"!--my teacher uses a different format for angles...)

    For the first question, break down the plane's airspeed and groundspeed into x and y components using these same formulas, then you can use

    airspeed_x + windspeed_x = groundspeed_x
    airspeed_y + windspeed_y = groundspeed_y

    to find the x and y components of the windspeed. Then use the last two formulas to convert back to speed and angle.

    EDIT: Don't bother with this, the way earboth did it is way easier.

    Does this make sense now? The basic idea is that airspeed, windspeed and groundspeed are all vectors, and that airspeed + windspeed = groundspeed. To add vectors, you break them down into x and y components, then combine the x with the x and the y with the y before converting back.
    Last edited by DavePercy; February 19th 2008 at 11:14 AM. Reason: used wrong angle measure
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  5. #5
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    Quote Originally Posted by iheartthemusic29 View Post
    I am soooo CONFUSED!

    1. An airplane is flying in the direction South 32 degrees East (S32E) with an airspeed of 860 kilometers per hour. Because of the wind, its groundspeed and direction are respectively 800 kilometers per hour and South 40 degrees East (S40E). Find the direction and speed of the wind.

    ...
    You are dealing with a triangle and the known parts of the triangle satisfy SAS.

    first side: 860 km (black)
    second side: 800 km (blue)
    included angle: 8

    Use Cosine rule to calculate the length of the 3rd side (red):

    d = \sqrt{860^2+800^2-2 \cdot 860 \cdot 8 \cdot \cos(8^\circ)} \approx 130.35 \ km (By the way that's a severe(?) gale)

    Now calculate the angle \alpha between the black and the red line. With this result it is easy to calculate the direction the wind is blowing:

    \cos(\alpha)=\frac{800^2-130.35^2 - 860^2}{-2 \cdot 130.35 \cdot 860} \approx 0.52002... ~\implies~ \alpha \approx 58.666^\circ

    Now calculate the direction from which the wind is blowing.

    (Did you get: 206.666 or S26.666W (?))
    Attached Thumbnails Attached Thumbnails vectors: HELP!!!!!-flugzeug_kurs.gif  
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  6. #6
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    Question The answer is....

    North 26.7 degrees East (N26.7E) and 130.35km/hour
    Now, the way my instructor told me to do it is:
    Since you have 32 degrees below the horizontal, you should add 270 to that so it'll be 302 degrees. She said call that vector p.
    p=<860cos302, 860sin302>
    Then, the other angle is 40 so you add that to 270 and that will be 310 degrees. She said that would be the resultant.
    r=<800cos310, 800sin310>
    So, she said the wind speed will be w=r-p.
    But your answers are correct. However, do you think you could explain it to me using the method she suggested? If you could, that would be a great help. Thanks!
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  7. #7
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    Question Hey...

    I figured it out!!!!!!!! Thanks!!!
    Now...if somebody can help me with the second problem, I would be forever grateful!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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  8. #8
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    Question So...

    I added some information to the second problem that I left out before.
    Here's what I got:
    airspeed= <580cos60, 580sin60>
    windspeed= <60cos45, 60sin45>
    Then I would add the airspeed and the windspeed to get the groundspeed.
    My answer came out to be 522.28 miles/hour.

    Can anybody confirm???
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  9. #9
    Newbie DavePercy's Avatar
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    Quote Originally Posted by iheartthemusic29 View Post
    airspeed= <580cos60, 580sin60>
    windspeed= <60cos45, 60sin45>
    Then I would add the airspeed and the windspeed to get the groundspeed.
    My answer came out to be 522.28 miles/hour.

    Can anybody confirm???
    If the airspeed angle is N60W, then the angle you use with the trig functions is 60+90 = 150, because N60W is 150 degrees counterclockwise from east.

    I'm not sure how you got the 522.28; I'm getting 567.44. I'd guess it was rounding error but it seems like a lot...
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  10. #10
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    Question I re-did it...

    and got the same thing you did. But what did you get for the angle?
    I used the tan^-1(y2+y1/x2+x1) and got -35.9 degrees. I then added 180 to that and got 144.1 degrees. Is that right?
    And once I know the angle, how can I tell what the bearing is (i.e., will it be N144.1W, S144.1W, etc.)?????
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  11. #11
    Newbie DavePercy's Avatar
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    Yes, -35.9 is in the SE quadrant, and the plane must be going in to NE quadrant because x2+x1<0 and y2+y1>0, so adding 180 gives you the angle counterclockwise from east. (Since tan^-1(y/x) = tan^-1(-y/-x) the angle will only ever be right or exactly 180 degrees off.) To turn it into a bearing subtract 90 to get N54.1W.
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