Asked by Lin
An airplane is traveling at a speed of 724 km per hour at a bearing of N30°E. If the wind velocity is 32 km per hour from the west, find the resultant speed and direction of the plane.
Answers
Answered by
kudum
Vp= 724(cos(60), sin(60))
= 724 (1/2 , (√3)/2)
= (362, 362√3)
Vw = 32(cos(0), sin(0))
= (32, 0)
VR = (√394^2 + (362√3)^2))
= 740.5 km/h
tanθ = (362√3)/394
θ = tan^-1(362√3)/394)
θ = 57.85 degree
N = 32.15 degree
= 724 (1/2 , (√3)/2)
= (362, 362√3)
Vw = 32(cos(0), sin(0))
= (32, 0)
VR = (√394^2 + (362√3)^2))
= 740.5 km/h
tanθ = (362√3)/394
θ = tan^-1(362√3)/394)
θ = 57.85 degree
N = 32.15 degree
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