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Relative velocity question from giancoli physics

  1. Aug 18, 2011 #1
    1. The problem statement, all variables and given/known data
    I attached the picture on the bottom (correction on the picture 30 degrees ---> 35 degrees sorry about that)
    Vpa: 600 km/h
    Vag:100 km/h

    and the question is to find the angle theta
    2. Relevant equations



    3. The attempt at a solution

    So I tried using Pythagorean theory by setting x and y

    [tex] (V_pa)^2 =y^2 + x^2 [/tex]

    set [tex] y = x/ (tan 35) [/tex]

    so then I solved x = 344.1458617 Km/h

    and since [tex] Sin/theta = (V_ag + x)/V_pa [/tex]

    [tex] /theta = 47.75271609 [/tex]

    sorry for ignoring the scientific notation,
    I am not used to it :(

    um anyways my answer was wrong
    because the "solution" chose to use the 'sine law'
    and that gives the right answer but I wanted to know what I did wrong to get the wrong answer


    much appreciated
     
    Last edited: Oct 2, 2011
  2. jcsd
  3. Aug 18, 2011 #2

    PeterO

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    Homework Helper

    Your diagram shows right angled triangles. The real situation may not be right angled, which would necessitate use of sin rule or cos rule
     
  4. Aug 18, 2011 #3
    Okay I will post the question up to see if it wasn't a right angle

    An airplane, whose air speed is 600km/h, is supposed to fly in a straight path 35 degrees north of east
    but a steady 100km/h wind is blowing from the north
    in what direction shuold the plane head?
     
  5. Aug 18, 2011 #4

    PeterO

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    Homework Helper

    [Responding to your original so I can copy your symbols]

    Your first line was incorrect

    [tex] (V_pa)^2 =y^2 + x^2 [/tex]

    It should read

    [tex] (V_pg)^2 =y^2 + x^2 [/tex]

    and since you don't know [tex] (V_pg)[/tex] it doesn't get you anywhere.

    You have to use the sine rule on the triangle including theta to solve
     
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