Results 1 to 3 of 3

Math Help - Bearing of wind

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    185
    Thanks
    1

    Bearing of wind

    A plane, whose speed in still air is 500 km/h, flies from a point A to a point B, 1560km due north of A. Because of the action of a constant wind, the plane must head in a direction whose bearing is 009 degrees. Given that the flight takes 3 hours,
    (i) Show that the speed of the wind is approximately 82.5km/h
    (ii) Find the bearing of the direction from which the wind is blowing.

    On another occasion the wind, whose speed is now 90km/h, is blowing from a direction whose bearing is 120 degrees. A second plane, whose speed in still air is 375 km/h, flies from A to a point C which is due east of A. Given that this flight also takes 3 hours.
    (iii) find the distance AC

    Part (i) is easy, but I have difficulty solving part (ii), can anyone help?
    Also, for part (iii), the bearing of the wind is 120 degrees, i can't draw the diagram, where is the angle 120 degrees??bearing of vector?
    I understand bearing of a fixed point from another fixed point, but bearing of a vector, i'm confused

    Thanks
    Last edited by acc100jt; September 21st 2007 at 07:12 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Are there no typos in your posted question?
    Bearing 009 degrees? What is that? Is that the same as bearing 9 degrees?

    Bearing 9 degrees is North 9deg East.
    Bearing is measured from the North axis, going clockwise.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    772
    Hello, acc100jt!

    A plane, whose speed in still air is 500 km/h,
    flies from a point A to a point B, 1560 km due north of A.
    Because of a constant wind, the plane must head in a direction whose bearing is 9.
    Given that the flight takes 3 hours,
    (i) Show that the speed of the wind is approximately 82.5 km/h
    (ii) Find the bearing of the direction from which the wind is blowing.
    Code:
        B *
          | *  3w
          |   *
          |     *
          |      * C
     1560 |     *
          |    *
          |   * 1500
          |9*
          | *
          |*
        A *
    The plane intends to fly from A to B: AB = 1560 km.

    It must fly to C, AC = 1500 km, due to the wind.

    Let w = wind speed (in km/hr).
    The wind blows from C to B in 3 hours: CB \,=\,3w

    Law of Cosines: . (3w)^2 \;=\;1560^2 + 1500^2 - 2(1560)(1500)\cos9^o \;=\;61218.56602

    Hence: . 3w \:=\:247.423859\quad\Rightarrow\quad w \:=\:82.47461968 \:\approx\:\boxed{82.5\text{ km/hr}}


    Law of Cosines: . \cos B \;=\;\frac{AB^2 + BC^2 - AC^2}{(AB)(BC)} \;=\;\frac{1560^2 + 247.5^2 - 1500^2}{2(1560)(247.5)} \;=\;0.951267483

    Hence: . B \:=\:71.513012931 \:\approx\:71.5^o

    Therefore, the wind is blowing from a direction
    . . with a bearing of 180^o - 72.5^o \:=\:\boxed{108.5^o}



    On another occasion the wind, whose speed is now 90 km/h,
    is blowing from a direction whose bearing is 120.
    A second plane, whose speed in still air is 375 km/h,
    flies from A to a point C, which is due east of A.
    Given that this flight also takes 3 hours.
    (iii) find the distance AC
    Code:
                                  D
                                  *     :
                              *    *    :
                   1125   *         *60:
                      *          270 *  :
                  *                   * :
              *                    30 *:
        A * * * * * * * * * * * * * * * * C
    The plane intends to fly from A to C.
    Due to the wind, it flies to D: . AD = 1125

    The wind blows from C to D: . CD = 270

    Law of Sines: . \frac{\sin A}{a} \:=\:\frac{\sin C}{c}\quad\Rightarrow\quad \sin A \:=\:\frac{270\sin30^o}{1125} \:=\:0.12

    Hence: . A \:=\:6.892102579 \:\approx\:6.9^o

    . . Then: . D \:=\:180^o - 30^o - 6.9^o \:=\:143.1^o


    Law of Sines: . \frac{d}{\sin D} \:=\:\frac{c}{\sin C} \quad\Rightarrow\quad d \:=\:\frac{1125\sin143.1^o}{\sin30^o} \:=\:1350.631451

    Therefore: . AC \;\approx\;\boxed{1350.6\text{ km}}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Speed of Wind
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: December 1st 2009, 08:15 AM
  2. Airplane Wind Speed?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 14th 2009, 03:15 AM
  3. Application - Airplane and Wind
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 10th 2009, 02:14 PM
  4. Algebra - Wind problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 31st 2007, 05:48 PM
  5. Speed of wind
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: January 10th 2007, 02:10 AM

Search Tags


/mathhelpforum @mathhelpforum