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Homework Help: Vectors Question

  1. Jan 18, 2006 #1
    A plane with a still air speed of 250 km/h is flying due NorthWest. At the same time a wind is blowing toward the South at 50 km/h. In what direction should the pilot head to continue travelling due North West? (The plane must continue with a 250 km/h velocity)

    My Answer:

    Velocity of plane (Vp) = 250
    Velocity of wind (Vw) = 50

    sin y = y/x = 50/250
    y = 11.5 degrees

    90 degrees - 11.5 degrees = 78.5 degrees

    The plane must head 78.5 degrees N of W to continue in a 45 degree
    N of W line.

    Is this correct? Any help is much appreciated.
    :smile:
     
  2. jcsd
  3. Jan 20, 2006 #2

    Tom Mattson

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    Your method looks wrong. You need to write down a vector for the velocity of the plane with respect to the air (call it [itex]\vec{v}_{PA}[/itex]) and a vector for the velocity of the air with respect to the Eart (call it [itex]\vec{v}_{AE}[/itex]. Then you add them up to get the velocity of the plane with respect to the Earth (call it [itex]v_{PE}[/itex]).

    [tex]\vec{v}_{PE}=\vec{v}_{PA}+\vec{v}_{AE}[/tex]

    From there, find your angle.
     
  4. Jan 20, 2006 #3

    lightgrav

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    You draw the vectors in the right directions, added tail-to-tip!
    your triangle doesn't have a 90-degree in it, but you can use
    the law of cosines since you do know two legs and the 135 angle.
     
  5. Jan 21, 2006 #4
    i think use the formula for resultsnt
    R=sq.root(p*p+Q*Q+2pqcos135)
    find the resultant.
    for direction use tan x=p cos 135/p+qsin135
     
  6. Jan 21, 2006 #5

    andrevdh

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    The resulting motion, [itex]\vec r[/itex], of the plane is the vector sum of is still air speed, [itex]\vec s[/itex], and the wind speed, [itex]\vec w[/itex]. The plane has to fly in the direction of the [itex]\vec s[/itex] vector in order to have a resultant motion in the direction of the [itex]\vec r[/itex] vector. Since the [itex]\vec r[/itex] vector is making an angle of [itex]45_o[/itex] with the "x-axis" its x- and y-components have the same magnitude. Assuming that the [itex]\vec s[/itex] vector makes an angle [itex]\theta[/itex] with the x-axis we can therefore say that:
    [tex]r_x\ =\ r_y[/tex]
    which gives
    [tex]s\cos(\theta)\ =\ s\sin(\theta)\ -\ 50[/tex]
     
    Last edited: Nov 29, 2006
  7. Jan 22, 2006 #6

    andrevdh

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    If my previous relation is a bit too challenging try solving for the angle between [itex]\vec r[/itex] and [itex]\vec s[/itex], say angle [itex]x[/itex], via the sine rule:
    [tex]\frac{\sin(x)}{w}=\frac{\sin(135^o)}{s}[/tex]
     
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