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NotTim

  • 3 years ago

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.

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  1. NotTim
    • 3 years ago
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    not physics

  2. LordHades
    • 3 years ago
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    Vector addition?

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

  4. LordHades
    • 3 years ago
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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.

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

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

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

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

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

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

  11. NotTim
    • 3 years ago
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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

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

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

  14. radar
    • 3 years ago
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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.

  15. radar
    • 3 years ago
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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

  16. NotTim
    • 3 years ago
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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?

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

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

  19. NotTim
    • 3 years ago
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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

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

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

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

  23. NotTim
    • 3 years ago
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    hello?

  24. NotTim
    • 3 years ago
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    computer went wonky.

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

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

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

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

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

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

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

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

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

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

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

  36. NotTim
    • 3 years ago
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    bump

  37. experimentX
    • 3 years ago
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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

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

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

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

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

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

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

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

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

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

  47. Dockworker
    • 3 years ago
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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

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

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

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

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

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

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

  54. Dockworker
    • 3 years ago
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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

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

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

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

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

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

  60. Dockworker
    • 3 years ago
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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

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

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

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

  64. NotTim
    • 3 years ago
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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.

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

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

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

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

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

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

  71. NotTim
    • 3 years ago
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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?

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

  73. NotTim
    • 3 years ago
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    135

  74. NotTim
    • 3 years ago
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    the wind blowing.

  75. Dockworker
    • 3 years ago
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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

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

  77. Dockworker
    • 3 years ago
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    315-180

  78. NotTim
    • 3 years ago
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    oohhhkay.

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

  80. Dockworker
    • 3 years ago
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    or a straight line

  81. NotTim
    • 3 years ago
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    ok.

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

  83. Dockworker
    • 3 years ago
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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

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

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

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

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

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

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

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

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

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

  93. NotTim
    • 3 years ago
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    yes.

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

  95. NotTim
    • 3 years ago
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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?

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

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

  98. Dockworker
    • 3 years ago
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    yw

  99. NotTim
    • 3 years ago
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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.

  100. radar
    • 3 years ago
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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.

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