2D Vector Addition | Airplane in the Wind | Boat Across the River

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  • Опубліковано 19 вер 2023
  • Look at three vector addition examples demonstrating how to calculate the vector sum of two vectors.
    Example 1
    A plane flies directly east at 30m/s through the air. Simultaneously the air blows to the south at 10m/s. Calculate the resultant speed of the airplane as well as its direction relative to East.
    Example 2
    Given the speed of the plane remains 30m/s and the wind remains 10m/s from the North, calculate the direction the airplane must point in order to travel directly East.
    Example 3
    Given the airplane is pointed 30 degrees North of East, and the wind is blowing 45 degrees South of East, in what direction and at what speed will the airplane travel relative to a stationary point.
    All three of these problems require breaking velocity vectors up into their components then creating sums of the components from each axis. Knowing the vector sums in each axis, we can use the resultant components to find the magnitude and direction of the resultant velocity vector.
    Note: It is possible to add many types of vectors together using the method outlined in this tutorial. Position, displacement, velocity, acceleration, force and momentum vectors come to mind...
    The topic of 2 dimensional (2D) vector addition comes up in introductory physics courses as well as high school math courses. These problems appear on high school physics exams, the AP Physics 1, AP Physics C Mechanics exams. They also show up in AP PreCalculus as well as on the A Level Physics and JEE Exams.

КОМЕНТАРІ • 3

  • @miketaylor3947
    @miketaylor3947 4 місяці тому

    What's the vector equation for wind correction angle and resultant ground speed in the last problem?
    Instead of computing drift angle and resultant ground speed as shown?
    I'm thinking we need three equations, yes?
    r = resultant, br = bearing, ps = plane speed, th = theta, ws = wind speed, wa = wind angle ; we are given ps, ws, wa, br, need to solve for th, mag r (rx, ry)
    rx = r sin br = ps sin th + ws sin wa
    ry = r cos br = ps cos th + ws cos wa
    ry/rx = tan br
    note: bearing angles clockwise from 0 degrees north
    OR, try to solve for R first?
    OR, don't use components and just use the geometry
    wca = - asin (ws/ps * sin (wa-br))
    gs = sqrt (ps^2+ws^2-2(ps)(ws)cos (br - wa + wca))
    Thanks!

    • @INTEGRALPHYSICS
      @INTEGRALPHYSICS  4 місяці тому

      See the 2nd example.

    • @miketaylor3947
      @miketaylor3947 4 місяці тому

      @@INTEGRALPHYSICS Thanks for your reply! Yes, I watched all three examples. My question was specific to example #3 where we don't get to enjoy the properties of a right triangle. I wrote down in my question what I thought, three unknowns, need three equations, then solve the linear algebra problem. Thanks in advance!