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Relative motion of a light plane

  1. Jun 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A light plane attains an airspeed of 470 km/h. The pilot sets out for a destination 840 km due north but discovers that the plane must be headed 24.0° east of due north to fly there directly. The plane arrives in 2.00 h. What were the (a) magnitude and (b) direction of the wind velocity? Give the direction as an angle relative to due west, where north of west is a positive angle, and south of west is a negative angle.

    2. Relevant equations
    (relative motion in two dimensions)
    Vpa= Vpb + Vba (the velocity of Vpa of P as measured by A is equal to the velocity Vpb of P as measured by B plus the velocity of Vba measured by A)

    3. The attempt at a solution
    Okay, so here was my valiant but wrong attempt. So first, I drew and labeled my diagram using the three following vectors: (1) Vpg: The plane's velocity in relation to the ground in which I had it angeled 24.0° east of due north (forms hypotenuse of triangle). (2) Vpw: The plane's velocity in relation to the wind in which I had it directly on the y-axis (forms adjacent of triangle ). And (3) Vwg: The wind's velocity in relation to the ground which I had angled westbound to cause the plane to go straight (forms opposite side of 24° angle in triangle). Next, I then made a relationship among the vectors using the triangle that I was able to draw: Vpg =Vpw + Vwg or Vwg= Vpg-Vpw. Where Vpg= 840km/2h= 0i + 420j (b/c only in y direction) and Vpw=470km/h*cos(24)i + 470km/h*sin(24)j= 429.366i + 191.166j. Therefore, Vwg=(0-429.366)i+(420-191.166)j= -429.66i + 228.833j. The magnitude is sqrt(429.66^2 + 228.833j^2) And the angle is tan^-1(191.166/429.366) +90 degrees. It seemed so right, where did I go wrong???
  2. jcsd
  3. Jun 3, 2007 #2

    Andrew Mason

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    You know the resultant velocity vector of the plane relative to land (420 km/hr due north). You know that this consists of the sum of two velocity vectors: the velocity of the air relative to land and the velocity of the plane relative to the air.

    You know the magnitude of the plane to air velocity vector (470 km/hr). You know its direction. You know that northern component of this velocity + the northern component of the air velocity relative to land adds up to 420 km/hr. Since the plane does not move east or west, you know that the sum of the east-west components of those two vectors is 0.

    From the above, set up two equations with two unknowns:

    [tex]\vec v_{pa}\sin (24^o) + \vec v_{al}\cos (x) = 0[/tex] (where x is the angle cclockwise from the due west direction).

    [tex]\vec v_{pa}\cos (24^o) + \vec v_{al}\sin (x) = 420[/tex]

    Solve for the two unknowns: [itex]|\vec v_{al}|[/itex] and the angle x.

  4. Jun 4, 2007 #3
    Thanks alot, that got me to the right answer
  5. May 31, 2008 #4
    please help me

    soo how do u solve the two equations i got stuck and what is the answer. i really need help.
  6. May 31, 2008 #5


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    Welcome to PF!

    Hi notsogood ! Welcome to PF! :smile:

    I take it you've got …
    Well, first tidy it up by putting the unknowns on one side:

    [tex]-\vec v_{pa}\sin (24^o)\,=\,\vec v_{al}\cos (x)[/tex]

    [tex]-\vec v_{pa}\cos (24^o)\,-\,420\,=\,\vec v_{al}\sin (x)[/tex]​

    And then … ? :smile:
  7. May 31, 2008 #6
    ok simultanious equations yee? well i only did general maths and i am in progress of learning 2 unit. soo b=xy but x and y are unkown soo my knowledge is limited to be able to solve equations which really gets on my nerves. i understand the problem but just cannot work it out. is there another way to solve it. another approach?
  8. Jun 1, 2008 #7


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    Hi notsogood! :smile:

    The only unknowns are val and x.

    There are two unknowns, and so you need to get rid of one of them.

    So …

    Hint: divide one equation by the other.

    Or square both equations, and then add them. :smile:
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