Vh = hor = 325cos340 + 40cos320,
Vh = 305 +30.6 = 335.6mph.
Vv = ver = 325sin340 + 40sin320,
Vv = -111.2 -25.7 = -136.9mph.
tanA = Vv/Vh = -136.9 / 335.6 = 0.4080,
A = - 22.2deg CW = 337.8 deg. CCW.
R = Vh / cosA = 335.6 / cos337.8 = 362.5mph @ 337.6 deg.
An airplane is flying on a compass heading (bearing) of 340 degrees at 325 mph. A wind is blowing with the bearing 320 degrees at 40 mph. Find the component form of the velocity of the airplane. then find the actual ground speed and direction of the plane
4 answers
Troll
cdsf
Airplane velocity A=(325 cos110, 325 sin110)
Wind velocity W=(40 cos130, 40 sin 130)
resultant = A+W = <-136.87, 336.04>
Actual ground speed = √(-136.87)^2 +(336.04)^2 = 362.85 mph
angle theta= arc tan (-336.04/136.87) = 112.16
bearing = 337.84
Wind velocity W=(40 cos130, 40 sin 130)
resultant = A+W = <-136.87, 336.04>
Actual ground speed = √(-136.87)^2 +(336.04)^2 = 362.85 mph
angle theta= arc tan (-336.04/136.87) = 112.16
bearing = 337.84
1 answer