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Relative motion of an airplane

  1. Feb 4, 2014 #1
    1. The problem statement, all variables and given/known data

    An airplane flies with a velocity of 55.0 m/s [ 35o N of W] with respect to the air (this is known as air speed). If the velocity of the airplane according to an observer on the ground is 40.0 m/s [53o N of W], what was the wind velocity?

    2. Relevant equations

    Pythagorean theorem, breaking into components, sine and cosine law

    3. The attempt at a solution

    I tried adding the negative of the airplane vector to the ground vector and using the pythagorean theorem to find the other side but apparently that is wrong. THE TEACHER POSTED A VIDEO TRYING TO EXPLAIN IT I STILL DIDNT UNDERSTAND IT. HERE IS THE VIDEO : http://www.educreations.com/lesson/view/module-1-2-question-2/16555555/
     
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  3. Feb 4, 2014 #2

    BvU

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    Hello Saeedawaw and welcome to PF. You've seen the template come by and for a first time user you're doing reasonably well in using it. One of the other conventions in PF is that we don't shout at each other, nor at the world in general. Upper case is considered shouting.

    I looked at the video and I now have great respect for your teacher. He acknowledges that it's not so easy and helps you on your way. So -- if you followed his suggestions -- you must have something already.
    Could you post the details of your calculation ? Because my impression is that what you intend to do is fully correct: if ##\vec {ground} - \vec {wind} = \vec {plane}## then ##\vec {ground} - \vec {plane} = \vec {wind}##. Pythagoras was proven right long ago, so what is it that doesn't work ?
     
  4. Feb 4, 2014 #3
    My bad. I meant to say "ps" but I don't really know how I ended up with caps haha. But when I do it I keep getting 21m/s in a 1degree angle in north of east. Is that right? And I thought I did it right
     
  5. Feb 5, 2014 #4

    BvU

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    Can't tell if it's right or not until you show the steps you take to get to these two numbers. My guess would be not, on the basis of a simple drawing. Your teacher made a drawing; did you ?
     
  6. Feb 5, 2014 #5

    BvU

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    Another thingy: don't believe everything. If you think you did it right, you can check that:

    If I state ground-wind=plane, scratch your head and find a simple example: I fly 100 m/s north wrt the air, the wind 40 m/s south wrt ground. Would my ground speed be 60 m/s north? I think it would. Now feed these numbers into your calculations and check if the outcome is correct!
     
  7. Feb 5, 2014 #6
    heres my work

    heres a picture of my work. Can you verify my steps? Im still in grade 10, but I took summer school and night school and im taking grade 12 physics online atm. It was stressing me out because no one of my classmates is helping until I found this website. image.jpg
     
  8. Feb 5, 2014 #7

    BvU

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    Your "should I do this" can be answered with no. Because you wanted ground - plane you want to do
    vw = (vw,x, vw,y ) = ( vg,x - va,x , vg,y - va,y )

    But I take it you did something like that and you got 21 m/s and 1 degree. It's perfect.
     
  9. Feb 5, 2014 #8
    I seriously can't thank you enough. Is there a way I can donate or support this forum? Im loving it here so far
     
  10. Feb 5, 2014 #9
    And on another note, i did the step you told me and got 15m/s for the magnitude, does that sound right?
     
  11. Feb 5, 2014 #10

    BvU

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    Oops, which magnitude was that ?
     
  12. Feb 5, 2014 #11
    The wind magnitude
     
  13. Feb 5, 2014 #12

    BvU

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    Don't understand. The 21 you found is the magnitude of the wind vector. What step brought you to 15 ?
     

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  14. Feb 6, 2014 #13
    I did it wrong, he just explained to me in a video. http://www.educreations.com/lesson/view/said-assignment-help/16688791/?ref=app [Broken] .

    In the video he said that in one of the component triangles the x is negative and the y is negative in the other one. Can you explain why is that? Thank you
     
    Last edited by a moderator: May 6, 2017
  15. Feb 6, 2014 #14

    BvU

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    I notice your teacher consistently draws the ground speed vector bigger than the airspeed vector. In your problem statement this is not the case. Could you check please?

    Marvelous that you can communicate with your teacher like this !
     
  16. Feb 6, 2014 #15

    BvU

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    Never mind. Teacher Bill also uses 55 and 40, and he admits it's not to scale.
    If you draw his red ground speed vector to scale and with the right angle, the blue one is going to point to 1 degree north of east.

    What you got was good. I don't have anything to bet on it, otherwise I would bet my reputation.

    Teacher's comment that sin and cos were interchanged: I would say you had ## \vec v_{a,x}## and ##\vec v_{a,y}## interchanged. Because just above you mention the y speed first too. Or is it a bit sloppy work ?
    Can't really read what's to the left of ##\vec v_{a,x}## and ##\vec v_{a,y}## on my cheapo screen....
     
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