Final answer:
The magnitude of the plane's resultant velocity is 53.4 m/s, with the direction being 13.0° south of east, calculated using the Pythagorean theorem and the tangent function.
Step-by-step explanation:
The student is asking to determine the magnitude and direction of the resultant velocity of a plane flying with a velocity of 52 m/s east, with a 12 m/s cross wind blowing the plane south. To find the resultant velocity, we can use the Pythagorean theorem because the vectors are perpendicular to each other.
The magnitude R can be calculated using the formula:
R = \sqrt{V_{plane}^2 + V_{wind}^2}
R = \sqrt{52^2 + 12^2} = \sqrt{2704 + 144} = \sqrt{2848} \approx 53.4 m/s
Next, to find the direction, we can calculate the angle θ with respect to the east direction using the tangent function:
θ = tan^{-1}(V_{wind}/V_{plane})
θ = tan^{-1}(12/52) \approx 13.0°
Because the wind blows the plane south, the direction of the resultant velocity is south of east.
Therefore, the magnitude of the plane's resultant velocity is 53.4 m/s, and the direction is 13.0° south of east.