An aircraft pilot wishes to fly from an airfield to a point lying s20e from the airfield . if there is a wind of 45 km/h from N80E and the cruising speed of the aircraft is 550km/h.
a) what direction should the pilot take?
b) what will her actual ground speed?
a) what direction should the pilot take?
b) what will her actual ground speed?
-
[ 550 * cos(θ) , 550 * sin(θ) ]
+
[ 45 * cos(90 - 80) , 45 * sin(90 - 80) ]
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[ 550cos(θ) + 45cos(10) , 550sin(θ) + 45sin(10) ] ===> (x , y)
S20E = 270 + 20 = 290
θ = tan^1(y/x)
tan(290) = y/x
sin(290)/cos(290) = y/x
sin(290) *x = y * cos(290)
sin(290) * ( 550cos(θ) + 45cos(10) ) = cos(290) * ( 550sin(θ) + 45sin(10) )
550sin(290)cos(θ) + 45sin(290)cos(10) = 550cos(290)sin(θ) + 45cos(290)sin(10)
550sin(290)cos(θ) - 550cos(290)sin(θ) = 45cos(290)sin(10) - 45sin(290)cos(10)
550sin(290 - θ) = 45sin(10 - 290)
550sin(290 - θ) = 45sin(-280)
sin(290 - θ) = (45/500)sin(-280)
290 - θ = sin^-1( (45/500)sin(-280) )
290 - θ ≈ 4.6
290 - θ ≈ 4.6
20 + 4.6 = 24.6
ground speed: √( (550cos(294.6) + 45cos(10))^2 + (550sin(294.6) + 45sin(10))^2) ≈ 540 km/h ====>294.6 and 10 are in rad NOT DEGREE
+
[ 45 * cos(90 - 80) , 45 * sin(90 - 80) ]
---------------------------------------…
[ 550cos(θ) + 45cos(10) , 550sin(θ) + 45sin(10) ] ===> (x , y)
S20E = 270 + 20 = 290
θ = tan^1(y/x)
tan(290) = y/x
sin(290)/cos(290) = y/x
sin(290) *x = y * cos(290)
sin(290) * ( 550cos(θ) + 45cos(10) ) = cos(290) * ( 550sin(θ) + 45sin(10) )
550sin(290)cos(θ) + 45sin(290)cos(10) = 550cos(290)sin(θ) + 45cos(290)sin(10)
550sin(290)cos(θ) - 550cos(290)sin(θ) = 45cos(290)sin(10) - 45sin(290)cos(10)
550sin(290 - θ) = 45sin(10 - 290)
550sin(290 - θ) = 45sin(-280)
sin(290 - θ) = (45/500)sin(-280)
290 - θ = sin^-1( (45/500)sin(-280) )
290 - θ ≈ 4.6
290 - θ ≈ 4.6
20 + 4.6 = 24.6
ground speed: √( (550cos(294.6) + 45cos(10))^2 + (550sin(294.6) + 45sin(10))^2) ≈ 540 km/h ====>294.6 and 10 are in rad NOT DEGREE