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Math Help - relative velocities

  1. #1
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    relative velocities

    A man was walking with a velocity of 4 i km/h when he obvserved that the wind was blowing from the north . When the man increased his velocity to 6 i km/h , he noticed that the wind seemed to be blowing from N 45 W . Find the speed and the direction of the wind .
    (Take the unit vector i to be in the direction due east.)
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  2. #2
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    Quote Originally Posted by hooke View Post
    A man was walking with a velocity of 4 i km/h when he obvserved that the wind was blowing from the north . When the man increased his velocity to 6 i km/h , he noticed that the wind seemed to be blowing from N 45 W . Find the speed and the direction of the wind .
    (Take the unit vector i to be in the direction due east.)
    If v is the man's velocity and w is the winds velocity, both relative to teh ground, then the wind's velocity, relative to him, is w- [/b]v[/b].

    Let the velocity of the wind (relative to the ground) be xi+ yj. The wind velocity relative to the man, when he is walking at 4i is xi+ yj- i= (x- 1)i+ j. The fact that the wind is from the north relative to him means x- 1= 0. When he is walking at 6i, the wind velocity, relative to him, is xi + y[/b]j[/b]- 6i= (x- 6)i+ yi. The fact that the wind is at N 45 W relative to him means that x- 6= y (legs of a isosceles right triangle).
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    Quote Originally Posted by HallsofIvy View Post
    If v is the man's velocity and w is the winds velocity, both relative to teh ground, then the wind's velocity, relative to him, is w- [/b]v[/b].

    Let the velocity of the wind (relative to the ground) be xi+ yj. The wind velocity relative to the man, when he is walking at 4i is xi+ yj- 4i= (x- 1)i+ j. The fact that the wind is from the north relative to him means x- 1= 0. When he is walking at 6i, the wind velocity, relative to him, is xi + y[/b]j[/b]- 6i= (x- 6)i+ yi. The fact that the wind is at N 45 W relative to him means that x- 6= y (legs of a isosceles right triangle).
    thanks , is the speed of wind the same for both cases ? And also , did you miss a 4 (in red) ?
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  4. #4
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    Hello, hooke!

    A fascinating problem . . . hope I got it right.


    A man was walking with a velocity of 4i km/h
    when he obvserved that the wind was blowing from the north.
    When the man increased his velocity to 6i km/h ,
    he noticed that the wind seemed to be blowing from N 45° W.
    Find the speed and the direction of the wind.
    Evidently, the wind is blowing from the northeast.

    The first phase looks like this:
    Code:
                      C
                    *
                  * |
              w *   |
              *     |
            * θ     |
        A * - - - - * B
               4

    The man walks from A to B\!:\;\;|\overrightarrow{AB}| \,=\,4
    The wind has vector \vec w \,=\,\overrightarrow{CA}\!:\;\;w \,=\,|\vec w|,\;\;\theta \,=\,\angle A
    The apparent wind vector is: . \overrightarrow{CB}

    We have: . \cos\theta \,=\,\frac{4}{w} \quad\Rightarrow\quad w\cos\theta \,=\,4 \quad\Rightarrow\quad w\sin\theta \,=\,\sqrt{w^2-16} .[1]


    The second phase looks like this:
    Code:
                    C
                    *
                  *   *
              w *       *
              *           *
            * θ         45° *
        A * - - - - - - - - - * D
                    6

    The man walks from A to D\!:\;|\overrightarrow{AD}| \,=\,4
    The wind vector \vec w is exactly the same.
    The apparent wind vector is: . \overrightarrow{CD},\;\angle D \,=\,45^o

    We have: . \angle C \:=\:180^o - \theta - 45^o \:=\:135^o - \theta


    Law of Sines: . \frac{w}{\sin45^o} \:=\:\frac{6}{\sin(135^o-\theta)} \quad\Rightarrow\quad w\sin(135^o-\theta) \:=\:6\sin45^o .[2]


    The left side is: . w\bigg[\sin135^o\cos\theta - \cos135^o\sin\theta\bigg] \;=\;w\bigg[\frac{1}{\sqrt{2}}\cos\theta - \left(-\frac{1}{\sqrt{2}}\right)\sin\theta\bigg]

    . . . . . . . . . . =\; w\bigg[\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta\bigg] \;=\; \frac{1}{\sqrt{2}}\,\bigg[w\cos\theta + w\sin\theta\bigg]

    . . Substitute [1]: . \frac{1}{\sqrt{2}}\left(4 + \sqrt{w^2-16}\right)


    Then [2] becomes: . \frac{1}{\sqrt{2}}\left(4 + \sqrt{w^2-16}\right) \:=\:6\left(\frac{1}{\sqrt{2}}\right) \quad\Rightarrow\quad 4 + \sqrt{w^2-16} \:=\:6

    . . . . . \sqrt{w^2-16} \:=\:2\quad\Rightarrow\quad w^2-16 \:=\:4 \quad\Rightarrow\quad w^2 \:=\:20 \quad\Rightarrow\quad w \:=\:2\sqrt{5}


    The wind speed is about: . \boxed{4.47\text{ km/hr}}


    Substitute into [1]: . w\cos\theta \:=\:4 \quad\Rightarrow\quad 2\sqrt{5}\cos\theta \:=\:4 \quad\Rightarrow\quad \cos\theta \:=\:\frac{2}{\sqrt{5}}

    . . Hence: . \theta \:\approx\:26.6^o


    The direction of the wind is: . \boxed{S\,63.4^o\,W}

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