Question #135433
A jet airliner, moving initially at 300 Km/h to the east, suddenly enters a region where the wind is blowing at 100 Km/h toward the direction 30.00 north of east. What is the new speed and direction of the aircraft relative to the ground?
1
Expert's answer
2020-10-15T10:36:56-0400


Solution

Given angle θ=720\theta=72^0

Speed of airliner

V=va2+vw2+2vavwcosθV=\sqrt{v_a^2+v_w^2+2v_av_w\cos\theta}

Where va,vwv_a, v_w are speed of airliner and wind with respect to ground.

Therefore new speed of airliner ith respect to ground can be expressed as

V=(300)2+(100)2+2×(300)×(100)cos30°V=\sqrt{(300)^2+(100)^2+2\times (300)\times (100)cos30^°}

V=389.82Km/hV=389.82Km/h


Direction

tanθ=vwva=100300=0.33tan\theta=\frac{v_w}{v_a}=\frac{100}{300}=0.33

θ=18.26°\theta= 18.26^°



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