The Magnitude of the resultant velocity is given by: Vr = squareroot of the addition of (2002 + 502) = 210m/s at 14.04degrees.
The velocity would be roughly 206.16 m/s at a heading of approximately if the airplane were simply following the wind and not attempting to course correct. A compass reading of 14.04 degrees (0 deg being North, 90 deg being East).
Y axis is north, and X axis is east. The line m=rise (y2-y1)/run (x2-x1) = 200/50 has a slope of 4/1. Since a2 + b2 = c2 on the triangle, 40,000 + 2,500 = 42,500. The hypotenuse of our right triangle, which represents the total distance traveled in one second, is equal to 206.16, which is the sq root of that number.
Now, we can determine the heading using TAN = Opposite / Adjacent (the angle x of the line on our graph). TAN (x) = 50/200 = 0.25 so x = 14.04 degrees, approximately.
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