Crosswind

A plane is traveling 45 degrees N of E at 320 km/h when it comes across a current from S of E at 115 degrees of 20 km/h. What are the airplane's new course and speed?

Final Answer:

Φ =  48.291 °
v =  327.3803 km/h

Step-by-step explanation:

$$\begin{gathered} \alpha =45{}^{\circ } \\ {v}_1=320\text{km/h} \\ \beta =115{}^{\circ } \\ {v}_2=20\text{km/h} \\ n=v_1\cdot \cos \alpha +v_2\cdot \cos \beta =v_1\cdot \cos 45°+v_2\cdot \cos 115°=320\cdot \cos 45°+20\cdot \cos 115°=320\cdot 0.707107+20\cdot (-0.422618)=217.8218 \\ e=v_1\cdot \sin \alpha +v_2\cdot \sin \beta =v_1\cdot \sin 45°+v_2\cdot \sin 115°=320\cdot \sin 45°+20\cdot \sin 115°=320\cdot 0.707107+20\cdot 0.906308=244.40033 \\ \tan \Phi =e:n \\ \Phi =\frac{180°}{\pi }\cdot \arctan (e/n)=\frac{180°}{\pi }\cdot \arctan (244.4003/217.8218)=48.291^{\circ }=48°17^{\prime }28" \end{gathered}$$

$$\begin{gathered} v^2=n^2+e^2 \\ v=\sqrt{n^2+e^2}=\sqrt{217.8218^2+244.4003^2}=327.3803\text{km/h} \end{gathered}$$

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