Results 1 to 4 of 4

Thread: relative velocities

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    107

    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.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,730
    Thanks
    3011
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    107
    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) ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    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 $\displaystyle A$ to $\displaystyle B\!:\;\;|\overrightarrow{AB}| \,=\,4$
    The wind has vector $\displaystyle \vec w \,=\,\overrightarrow{CA}\!:\;\;w \,=\,|\vec w|,\;\;\theta \,=\,\angle A $
    The apparent wind vector is: .$\displaystyle \overrightarrow{CB}$

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


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

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

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


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


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

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

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


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

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


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


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

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Velocities help
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Oct 4th 2009, 10:36 PM
  2. Relative Velocities
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: Aug 22nd 2009, 06:48 AM
  3. Velocities of curves.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 18th 2009, 06:42 AM
  4. Distances and velocities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Mar 17th 2009, 01:23 PM
  5. relative velocities
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Sep 16th 2008, 09:00 PM

Search Tags


/mathhelpforum @mathhelpforum