The displacement vector for a 15.0 second interval of a jet airplane's flight is (3150, −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?
2006-11-13 14:27:10 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
(sqrt(3150^2+2430^2))/15
265.2244 m/s
arctan(y/x)
arctan(-2430/3150)
-37.647 degrees
2006-11-13 14:30:32 · answer #1 · answered by nobody722 3 · 0⤊ 0⤋
Well, lets say 3150 meters in the x direction, and -2430 in the y direction.
Vx = 3150/15 meters/second
Vy = -2430/15 meters /second
Vavg = sqrt ( Vx^2 +Vy^2)
Angle = arctan ( 2430/3150) angle is below the x axis
measured with positive x axis = 0 counterclockwise the angle would be 360 - arctan(2430/3150)
2006-11-13 22:35:00 · answer #2 · answered by Roadkill 6 · 0⤊ 0⤋
You would not ask this question if you did not already know the answer. Please dazzle us all with your answer on the add comments line of your question.
2006-11-13 22:42:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋