The displacement vector for a 15.0 second interval of a jet airplane's flight is (3850, −2430) m.?
The displacement vector for a 15.0 second interval of a jet airplane's flight is (3850, −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?
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Verified answer
s = √(3850² + (-2430)²)m/15 sec.
θ = arctan(3850/(-2430))
Get the value of the vector, D = ?(2250 ² + 2430 ²) divide that by 15 seconds to get velocity in m/s perspective you will get by arctan(?2430/2250) draw a diagram to confirm of the perspective. .