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Hi, need help with vectors in diff directions

  1. Jan 3, 2007 #1
    1. The problem statement, all variables and given/known data

    An airplane starting from airport A flies 247 km east, then 281 km at 29.3 west of north, and then 385 km north to arrive finally at airport B. The next day, another plane flies directly from A to B in a straight line.

    In what direction should the pilot travel in this direct flight? Use the counter-clockwise as the positive angular direction from due east, between the limites of -180 and 180 degrees. Assume there is no wind during these flights. Answer in degrees.

    part 2
    HOw far will the pilot travel?

    2. Relevant equations

    Now i looked in my book and found this hefty equation/law:
    r^2 = A^2 + B^2 -2ABCosTheta
    I used this and found the answer to the second part, which is 639.493 km (I'm right because i checked it online)

    Now I'm probably lost because i did the second part before the first part.
    I get confused with direction involving angles because sometimes it's confusing if I have to use a negative sign or not, etc etc.

    3. The attempt at a solution

    This problem is hard to do it on the comp, but i did use the law twice in order to attain the final vector.

    Can you get the angle direction? If i do find it, i will post the answer with directions.
    Thanks :)

    well i tried and used up all my possible tries so i got 0 pts on it but whatever, correct answer was 80.1422
    Last edited: Jan 3, 2007
  2. jcsd
  3. Jan 3, 2007 #2


    User Avatar

    Staff: Mentor

    Do you know how to convert between polar and rectangular coordinates for vectors? You are given vectors essentially in polar notation (magnitude, direction), but it's a lot easier to add vectors by adding their components in rectangular notation: (x, y) or (E, N) in your case. The easiest way to do this problem is convert each of the initial flight vectors into their rectangular form, add the two sets of components to get the overall resultant, and then convert that answer back to the polar format that the question wants its answers in. Does that make sense?
  4. Jan 3, 2007 #3
    haha i'm sorry to say that no it doesn't.
    Actually one of my answers was 80.14, yet i didn't put that because of my fear of getting the sign wrong (and i had already used up my choices)

    I actually redrew my diagram and made it somewhat scale.
    I basically had a scalene triangle in which the big/inner angle was 155 and i needed to look for the top angle, which was 9.98ish.
    I did 180-(90+9.8) and got that 80.14 answer, but just my bad luck and stupid answering that i got it wrong all those times.
    It's confusing because it said something about the counterclockwise from east being positive.. but.. bah


    i kept looking to find the right side of the angle, and kept slapping a negative..
    when i should have been finding the left angle shaded, and positive.
    dangit.. makes me mad.

    thanks for your time :)
  5. Jan 3, 2007 #4
    Since you already submitted the problem maybe I can help you understand what's going on and how you can figure out this kind of thing in the future. It's actually really easy once you know what's going on.
    It's much better if you understand what is happening in the equation and how we get the equation.

    http://www.math.hmc.edu/calculus/tutorials/reviewtriglogexp/right_triangle.gif [Broken]
    (from http://www.math.hmc.edu/calculus/tutorials/)

    You should remember that you can break up any vector into its X and Y components, and when you add those two vectors together you get the original vector back. Think of a right triangle. The hypotenuse is original vector (R), the horizontal side of the triangle is the X component of the vector, and the vertical side is the Y component. This is an EXTREMELY important principle that you will use over and over again!

    Remember these trigonometric rules (SohCahToa):
    sin \theta = \frac{y}{R}

    cos \theta = \frac{x}{R}
    where R is the length of the resultant vector, x is the length of the horizontal component and y is the length of the vertical component (as in the figure).

    Rearranging the above equations, you can find the horizontal and vertical components using:
    y = R sin \theta

    x = R cos \theta

    Break up all the vectors into their X and Y components (you won't have to do this with the vectors pointing east and north because they aren't at an angle to the X or Y axis), add all the X components together and call it [tex]R_x[/tex], add all the Y components together and call it [tex]R_y[/tex].

    You now have the resultant of the vector from point A to B.

    To find the angle of the resultant, you can use the Toa part:
    tan \theta = \frac{y}{x}
    \theta = tan^{-1}(\frac{y}{x})

    Plug your numbers in for the total X and total Y components of the final resultant vector and you have your angle.

    I hope this helps you understand where the hefty equation comes from. Usually (if the textbook is any good) an explanation will accompany the equation, showing you how the equation is derived. If you memorize a bunch of equations, you don't understand physics. Try to understand!
    Last edited by a moderator: May 2, 2017
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