The displacement vector for a15.0 second interval of a jet airplane\'s flight is

Question

The displacement vector for a15.0 second interval of a jet airplane's flight is (2.15e+3, 2430) m. (a) What is the magnitude of theaverage velocity? (b) At what angle, measured from the positivex axis, did the airplane fly during this timeinterval?
(a) m/s
(b) (a) m/s
(b) The displacement vector for a15.0 second interval of a jet airplane's flight is (2.15e+3, 2430) m. (a) What is the magnitude of theaverage velocity? (b) At what angle, measured from the positivex axis, did the airplane fly during this timeinterval?
(a) m/s
(b)

Explanation / Answer

The displacement vector for a 15.0 second interval of a jetairplane's flight is ( 2.15e+3, 2430) m a) Now vx = (2.15*103m) / (15.0s)             = 0.143*103m/s          vy= (-2430m) / (15.0s)              =-162 m/s Now the magnitude of the average velocity is                 v = [vx2 +vy2] b) Direction is = tan-1(vy /vx)                     = tan-1[(-162m/s) / (143.3m/s)]                     = 138.5o with respect to the + x axis

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