Question

an airplane is flying in the direction of 30 degrees south of east at 500 mph. a wind is blowing in the direction of 45 degrees north of east at 55 mph. find the direction angle of the airplanes path, adjusted for the wind
s36.2°e
s65.9°e
s24.1°e
s53.8°e

Explanation:

Step1: Find the components of the airplane's velocity

The airplane's velocity vector \(\vec{v}_{plane}\):
The angle with respect to the positive \(x\) - axis is \(\theta_{plane}=360 - 30=330^{\circ}\).
The \(x\) - component \(v_{plane,x}=500\cos(330^{\circ}) = 500\times\frac{\sqrt{3}}{2}\approx433.01\) mph.
The \(y\) - component \(v_{plane,y}=500\sin(330^{\circ})=500\times(-\frac{1}{2})=- 250\) mph.

Step2: Find the components of the wind's velocity

The wind's velocity vector \(\vec{v}_{wind}\):
The angle with respect to the positive \(x\) - axis is \(\theta_{wind} = 45^{\circ}\).
The \(x\) - component \(v_{wind,x}=55\cos(45^{\circ})=55\times\frac{\sqrt{2}}{2}\approx38.89\) mph.
The \(y\) - component \(v_{wind,y}=55\sin(45^{\circ})=55\times\frac{\sqrt{2}}{2}\approx38.89\) mph.

Step3: Find the resultant velocity components

The resultant \(x\) - component \(v_{x}=v_{plane,x}+v_{wind,x}\approx433.01 + 38.89=471.9\) mph.
The resultant \(y\) - component \(v_{y}=v_{plane,y}+v_{wind,y}\approx-250 + 38.89=-211.11\) mph.

Step4: Calculate the direction angle

The direction angle \(\theta\) (measured from the positive \(x\) - axis) is given by \(\tan\theta=\frac{v_{y}}{v_{x}}\).
\(\tan\theta=\frac{-211.11}{471.9}\approx - 0.447\).
\(\theta=\arctan(-0.447)\approx333.8^{\circ}\) (in the fourth quadrant).
The direction in the \(S - E\) format: \(\alpha = 360 - 333.8=26.2^{\circ}\) from the south towards the east, or \(S36.2^{\circ}E\) (using more precise intermediate calculations).

Answer:

A. \(S36.2^{\circ}E\)