Asked by Arielle
A plane flies with a heading of 48 degrees NW and an air speed of 584 km/h. It is driven from its course by a wind of 58.0 km/h from 12.0 degrees SE. Find the ground speed and the drift angle of the plane.
Answers
Answered by
Henry
Vp' = Vp + Vw,
Vp'=584km/h @ 138deg + 58km/h @ 348deg.
X = hor. = 584cos138 + 58cos348,
X = hor. = -434 + 56.73 = -377km/h.
Y = ver. = 584sin138 + 58sin348,
Y = ver. = 391 + (-12) = 379km/h.
tanA = Y/X = 379 / -377 = -1.0051.
A = -45deg.,CW.
A = -45 + 360 = 315deg.,CCW.
Vp' = sqrt((-377)^2+(379)^2 = 535km/h = Velocity of plane.
Vp'=584km/h @ 138deg + 58km/h @ 348deg.
X = hor. = 584cos138 + 58cos348,
X = hor. = -434 + 56.73 = -377km/h.
Y = ver. = 584sin138 + 58sin348,
Y = ver. = 391 + (-12) = 379km/h.
tanA = Y/X = 379 / -377 = -1.0051.
A = -45deg.,CW.
A = -45 + 360 = 315deg.,CCW.
Vp' = sqrt((-377)^2+(379)^2 = 535km/h = Velocity of plane.
Answered by
Henry
Correction:
A = -45 + 180 = 135deg., CCW. Q2.
Vp' = 535km/h @ 135deg.
A = -45 + 180 = 135deg., CCW. Q2.
Vp' = 535km/h @ 135deg.
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