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Answer to Question #1243 in Mechanics | Relativity for Jessica

Question #1243
The displacement vector for a 15.0 second interval of a jet airplane's flight is (2850, −2430) m. (a) What is the magnitude of the average velocity? (b) At what angle, measured from the positive x axis, did the airplane fly during this time interval? Express the angle as a number between −180° and +180°
Expert's answer
The total displacement is
S = √ (28502 + 24302) = 3745.32 m
a) The average velocity : V = S/t =3745.32 / 15 = 249.69 m/s
b) tan (α) = -2430/2850 = 0.853
α = arctan (-0.853) = - 40.45°

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