Answer to Question #1243 in Mechanics | Relativity for Jessica

Question

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°

Answer 1

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°