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NotTim

A small aircraft, on a heading of 225 degrees, is cruising at 150km/h. IT is encountering a wind blowing from a bearing of 315 degrees at 35km/h. Determine the aircraft's ground velocity.

  • 2 years ago
  • 2 years ago

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  1. NotTim
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    not physics

    • 2 years ago
  2. LordHades
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    Vector addition?

    • 2 years ago
  3. NotTim
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    application of vector addition, yah.

    • 2 years ago
  4. LordHades
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    Draw it out. Find it's components. Add their x componants and y componants separately. Put them back together in a triangle.

    • 2 years ago
  5. NotTim
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    i somehow figured out how to get resultant, but not the angle...

    • 2 years ago
  6. NotTim
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    im not suppose to use component method

    • 2 years ago
  7. LordHades
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    Use tan(y/x)

    • 2 years ago
  8. LordHades
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    Oh I see. THen I'm afraid I can be of no help to you.

    • 2 years ago
  9. NotTim
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    ok...oh well. thx u

    • 2 years ago
  10. radar
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    draw it out, and remember when directions are concerned, 0 degrees is North.

    • 2 years ago
  11. NotTim
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    i did....but i didnt draw it here. i got the magnitude, but not the angle. i think though that i might have read the answer wrong...im still double checking again. but thank you

    • 2 years ago
  12. NotTim
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    yeah, i forgot it was bearings

    • 2 years ago
  13. radar
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    O.K, good luck on this problem it is a good question

    • 2 years ago
  14. radar
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    Oh yes, note that the wind is FROM a bearing of 315 degrees in other words the force is towards E 45 S.

    • 2 years ago
  15. radar
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    Note also the angle between the aircraft and the wind vector is 90 degrees. as aircraft is flying on a bearing of W 45 S

    • 2 years ago
  16. NotTim
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    hey, radar, i got 13.1 degrees (not considering bearings), as the angle between the resultant and the aircraft's cruising speed. Um. What should I do with that?

    • 2 years ago
  17. NotTim
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    @Radar Hey....do you have time? I need you to please revisit this...

    • 2 years ago
  18. NotTim
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    |dw:1333520833003:dw|

    • 2 years ago
  19. NotTim
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    thats how i came to my angle. the correct answer is a BEARING of 211.9 degreees. my answer is not a bearing

    • 2 years ago
  20. lexi_rocks_lexi
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    ur question is a hard 1. i tried 8 times and cant figure it out. sorry

    • 2 years ago
  21. NotTim
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    ok. its alright.

    • 2 years ago
  22. bobobobobb
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    Okay ive done this before. Let me dig up my notes.

    • 2 years ago
  23. NotTim
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    hello?

    • 2 years ago
  24. NotTim
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    computer went wonky.

    • 2 years ago
  25. bobobobobb
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    Sorry. Were you given the answer?

    • 2 years ago
  26. NotTim
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    yeah.its 154.0km/h at a bearing of 211.9 degrees.

    • 2 years ago
  27. bobobobobb
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    Draw your version of the diagram. I wanna see if we are using the same one

    • 2 years ago
  28. NotTim
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    i drew 2...

    • 2 years ago
  29. NotTim
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    or do you want what the txtbook has>?

    • 2 years ago
  30. bobobobobb
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    yeah that would be good

    • 2 years ago
  31. NotTim
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    |dw:1333522792176:dw|

    • 2 years ago
  32. Hero
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    I like how "not physics" is the first thing NotTim says.

    • 2 years ago
  33. NotTim
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    you know, just in case.

    • 2 years ago
  34. bobobobobb
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    So we are finding the resultant vector?

    • 2 years ago
  35. NotTim
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    i already found it (somehow...I don't really know how). I'm looking for the angle now.

    • 2 years ago
  36. NotTim
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    bump

    • 2 years ago
  37. experimentX
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    find out the individual horizontal and vertical factors ... add them or substract them according to necessity ... and reconstruct them. this is how these kinds of problems are done

    • 2 years ago
  38. NotTim
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    argghh. that's component method. im not suppose to use component method!

    • 2 years ago
  39. experimentX
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    since they don't have same directions, i don't think you have other choice ...

    • 2 years ago
  40. NotTim
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    what about application of vector additions? using sin adn cos law?

    • 2 years ago
  41. experimentX
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    let's try it.

    • 2 years ago
  42. Dockworker
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    \[\tan(\theta)=\frac{35}{150}, \theta=13.13\deg\] subtract from 225

    • 2 years ago
  43. NotTim
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    r^2=150^2+35^2-2(150)(35)cos90

    • 2 years ago
  44. Dockworker
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    theta is angle between resultant and original flight

    • 2 years ago
  45. NotTim
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    wait...so trig ratios work, even with non-right angle triangles?

    • 2 years ago
  46. experimentX
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    |dw:1333520171086:dw|

    • 2 years ago
  47. Dockworker
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    it is a right triangle because the plane is flying at 225 degrees, wind coming from 315 degrees or going at a bearing of 135

    • 2 years ago
  48. experimentX
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    |dw:1333520250197:dw|

    • 2 years ago
  49. Dockworker
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    225-135=90

    • 2 years ago
  50. NotTim
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    its 315...

    • 2 years ago
  51. NotTim
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    im sure thats not an error in my grammar

    • 2 years ago
  52. Dockworker
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    blowing from a bearing of 315º is the same as blowing at a bearing of 135º

    • 2 years ago
  53. experimentX
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    |dw:1333520364868:dw|

    • 2 years ago
  54. Dockworker
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    if there was wind ahead of you at a bearing of 315 degrees and it passed by you, it'd be going at 135º

    • 2 years ago
  55. NotTim
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    dock, i dont udnerstand. how would i derive that info?

    • 2 years ago
  56. experimentX
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    i think that would be 180 - (315-225) and apply cosine law .. i think that might work

    • 2 years ago
  57. Dockworker
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    if i tell you wind is coming from in front of you, which way is it going to push you?

    • 2 years ago
  58. Dockworker
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    the same way its coming from?

    • 2 years ago
  59. NotTim
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    no; I know the different of to and from (Radar explained that); how would I figure out:|dw:1333531452863:dw|

    • 2 years ago
  60. Dockworker
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    you have a magnitutude of 35 km/h wind, you shift the vector over to the end of the other vector

    • 2 years ago
  61. NotTim
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    yeah, but how EXACTLY do i figure out the new angle/bearing formed is 135?

    • 2 years ago
  62. Dockworker
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    |dw:1333520788256:dw| you could do as you did above.

    • 2 years ago
  63. Dockworker
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    draw a line from 315 bearing through the origin

    • 2 years ago
  64. NotTim
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    Yes, I understand that. I want to know how I can derive 135 from 315 when i get the equilibrant vector, like you did.

    • 2 years ago
  65. Dockworker
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    its just intuitive, not really something you should have to prove

    • 2 years ago
  66. Dockworker
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    just the wording of the problem

    • 2 years ago
  67. NotTim
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    no, but how do i figure that out myself? to get 135 from the equilibrant of 315?

    • 2 years ago
  68. NotTim
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    i dont know if im making sense here...

    • 2 years ago
  69. Dockworker
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    they aren't giving you the true bearing of the wind's velocity

    • 2 years ago
  70. Dockworker
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    they are giving it to you from a different reference

    • 2 years ago
  71. NotTim
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    I understand that. Yes. Very much so. The difference between to and from. But the angle, how did you derive your angle like that?

    • 2 years ago
  72. Dockworker
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    its so far up now. what angle did i derive?

    • 2 years ago
  73. NotTim
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    135

    • 2 years ago
  74. NotTim
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    the wind blowing.

    • 2 years ago
  75. Dockworker
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    its just a reference frame. from an observer's point of view, the wind is at a bearing of 315º. from the wind's perspective, the observer is at abearing of 135º

    • 2 years ago
  76. NotTim
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    How did you calculate 135degrees?

    • 2 years ago
  77. Dockworker
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    315-180

    • 2 years ago
  78. NotTim
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    oohhhkay.

    • 2 years ago
  79. Dockworker
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    the observer (plane) and wind, if looking at each other, form a 180º line of sight

    • 2 years ago
  80. Dockworker
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    or a straight line

    • 2 years ago
  81. NotTim
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    ok.

    • 2 years ago
  82. NotTim
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    now that i figured that out, what's our next step?

    • 2 years ago
  83. Dockworker
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    you know the magnitude of the plane's velocity and window's velocity. they form a right angle. the picture i drew above shows this. use arctan

    • 2 years ago
  84. NotTim
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    we werent taught arctan (to add to my problems)

    • 2 years ago
  85. NotTim
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    or, that was long before

    • 2 years ago
  86. Dockworker
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    this will give you the angle between the resultant and the plane's direction

    • 2 years ago
  87. NotTim
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    oh. trig ratios. tanangle=o/a right

    • 2 years ago
  88. Dockworker
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    arctan(35/150) is approximately 13.13º

    • 2 years ago
  89. NotTim
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    how do i know which one is my opposite and adjacent?

    • 2 years ago
  90. Dockworker
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    |dw:1333521664877:dw|

    • 2 years ago
  91. NotTim
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    oh bugger. I thought we were looking for another angle. this clarfies much,.

    • 2 years ago
  92. Dockworker
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    as you can see, the magnitutde can also be calculated with pythagorean theorem

    • 2 years ago
  93. NotTim
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    yes.

    • 2 years ago
  94. NotTim
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    i was just seeing if i could do this if it werent a right angle triangle. apparently i couldnt.

    • 2 years ago
  95. NotTim
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    ok. we're near the end now. how do i figure out hte angle, as in bearings? we've got 13.13 degrees. wat next?

    • 2 years ago
  96. NotTim
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    thx with sticking with me here folks, greatly appreciated.

    • 2 years ago
  97. Dockworker
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    subtract that angle from the plane's bearing

    • 2 years ago
  98. Dockworker
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    yw

    • 2 years ago
  99. NotTim
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    thank you all. your assistance has been greatly appreciated. I wish you the best of luck in your studies, and that you are able to help others as greatly as you have done for me.

    • 2 years ago
  100. radar
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    Had to leave, but I see that you got some excellent help and advice from Dockworker. I agree with the results. That angle of 13.13 degrees is not the bearing but it is the correct deviation from his original bearing. The bearing rose goes clockwise with 0 degrees north 180 degrees due south etc.

    • 2 years ago
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