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Homework Help: Relative velocity

  1. Oct 28, 2005 #1
    Pretty easy relative velocity question but I suck at Physics. I wanted to check my diagram and if it is not right than ask for help on a mathematical aproach because mine is not working.
    The question is: An airplane, who air speed is 600 km/h, is supposed to fly in a straight path 35.0 degrees north of east. But a steady 100 km/h wind is blowing from the north. In what direction should the plane head?
    http://www.oddworldz.com/octachoron/qyestuib53.jpg [Broken]
    It's not artistic but I hope it's understandable. The angle between the black arrow from origin and either vertical or horizontal is what I need to find. By the way: not to scale.
    Answer at back of book is: 42.9 degrees N of E
    This is the beginning of my approach which probably doesn't follow all math rules and is therefore ridiculous.
    600cosA = xcos35
    600sinA - 100 = xsin35
    Thanks for any help or hints or links on how to do math or anything.
    If no one answers before 7:00AM PST tomorrow don't bother because I'll check the solutions book at school.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Oct 28, 2005 #2
    You want to use vector addition, you have th emagnitude of the plane's velocity as 600km/h , and the total desired velocity vector (direction), you'll want to find a combination of three vectors (plane, total, and air) such that the sum of the plane vector and air vector = the total vector.
  4. Oct 28, 2005 #3
    OK. I have no idea what that meant. It seems like a convuluted re-statement of the question to be honest.
    http://img470.imageshack.us/img470/1309/redone6wk.th.jpg [Broken]
    I tried to show how I broke them down into components here.
    In Y:
    -100 for the wind
    600sinA for breakdown of plane velocity in Y
    In X
    600cosA for plane v
    that's where I got my approach from.
    Sooo.... can anybody please help me out.
    Last edited by a moderator: May 2, 2017
  5. Oct 28, 2005 #4
    Can't help you cause we can't see your pictures..
  6. Oct 28, 2005 #5
    Last edited by a moderator: May 2, 2017
  7. Oct 28, 2005 #6
    Try setting up a system of equations for the X and Y components, you should have 2 equations and 2 unknowns.
  8. Oct 28, 2005 #7
    Refering to my first entry:

    600cosA = xcos35
    600sinA - 100 = xsin35

    If you are saying that it is correct so far and those equations are right than I simply don't have the mathematical skills to solve it, of course I am assuming you are an authority figure on the area of Physics because you have 2000 posts.

    So if all that's right just confirm and I'll fool around with them for a couple more hours.
  9. Oct 28, 2005 #8
    I'm no expert, this could be wrong. I'm on medication so I'm not thinking straight, but no one else is helping you..

    [itex] \cos(x) = \sin(90 - x) [/itex]. I would try applying this to the right hand side of the first equation, might make things easier for you.
  10. Oct 28, 2005 #9
    I am aware of that and I wasn't complaining. Maybe I should though because now I learned you're high. I'm only kidding, sorry you have a disease or mental problem or something.

    Thanks for the help, this is going to take me awhile. It took me 15 minutes to figure out what that last thing you said meant.
  11. Oct 28, 2005 #10
    I broke my arm and am on prescription pain killers, I'm mentally intact thanks :)
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