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Homework Help: Calculating resultant velocity

  1. Oct 22, 2012 #1
    An airplane has an air speed of 4.20 × 10^2 km/h [N 45° E]. The wind speed is 30 km/h
    to the west.

    a) What is the airplane’s resultant velocity?

    I decided to use the trigonometric solution, but I am confused on how to break the equation down in order to get 399 km/h

    c2 = a2 + b2 - 2ab.cosC
    c2 = (4.20 x 10^2)^2 + (30)^2 - 2( 420) x (30)cos135
    c2 = (17640000 + 900) - 17819
    c2 = 17623081
    c2 = ??? (I tried square root of 17623081, but of course this is far off from the answer)

    Please assist!
  2. jcsd
  3. Oct 22, 2012 #2
    You're on the right track with the trigonometric approach, but you went about it wrong. The initial velocity you have is 4.2 x 10^2 at an angle of 45 degrees North of East. Isn't your c term the 4.2 x 10^2?
    From here you have a wind slowing down the x direction of the plane.
    The velocity you have now is incorporating both the x and y directions. How can you divide the velocity up so you can deal with the x and y directions separately?
  4. Oct 22, 2012 #3
    Hmm.. I am not sure what you mean. I am terrible at this! I don`t know the velocity (the 339 km/h is the answer I am suppose to find) so how would I divide it up?
  5. Oct 22, 2012 #4
  6. Oct 22, 2012 #5
    Yes, but I am not sure if I am doing it correctly.
    Would a = 420 km/h [N 45 E] + b = 30 km/h [W] equals the resultant? But, when drawing it, at what angle would 30 km/h [W] be?
  7. Oct 22, 2012 #6
    Draw it.

    You will see it's a simple case of a 30 km/h headwind at 45 degrees off the plane's nose.

    So the effect of the headwind along the path of the plane will be 30*cos (45). Subtract that from 420 and you have your answer.

    You can use trig if you like - but a bit of thought will often save you having to remember complicated formulae.
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