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 = √ (2850^{2} + 2430^{2}) = 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°

S = √ (2850

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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