Asked by Anonymous
An airplane has a velocity relative to the air of 181 m/s in a westerly direction. If the wind has a speed relative to the ground of 53 m/s directed to the north, what is the speed of the plane relative to the ground?
Answers
Answered by
Henry
Vpg = Vpa + Vag
Vpa = -181m/s @ 180deg.
Vag = 53m/s @ 90deg.
V(hor) = -181cos180 + 53cos90,
V(hor) = 181 + 0 = 181m/s.
V(ver) = -181sin180 + 53sin90,
V(ver) = 0 + 53 = 53m/s.
V^2 = X^2 + Y^2,
V^2 = (181)^2 + (53)^2,
V^2 = 32761 + 2809 = 35570,
V=sqrt(35570) = 189m/s=Speed of plane relative to ground.
Vpa = -181m/s @ 180deg.
Vag = 53m/s @ 90deg.
V(hor) = -181cos180 + 53cos90,
V(hor) = 181 + 0 = 181m/s.
V(ver) = -181sin180 + 53sin90,
V(ver) = 0 + 53 = 53m/s.
V^2 = X^2 + Y^2,
V^2 = (181)^2 + (53)^2,
V^2 = 32761 + 2809 = 35570,
V=sqrt(35570) = 189m/s=Speed of plane relative to ground.
Answered by
Henry
DEFINITIONS
Vpg = Speed of plane relative to ground.
Vpa = Speed of plane relative to air.
Vag = Speed of air relative to ground.
Vpg = Speed of plane relative to ground.
Vpa = Speed of plane relative to air.
Vag = Speed of air relative to ground.
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