Question

Solve each problem. See Examples 5 and 6.

Bearing and Ground Speed of a Plane An airline route from San Francisco to Honolulu is on a bearing of 233.0°. A jet flying at 450 mph on that bearing encounters a wind blowing at 39.0 mph from a direction of 114.0°. Find the resulting bearing and ground speed of the plane.

Accepted Answer

Identify the vectors involved: the plane's velocity vector and the wind's velocity vector. The plane's velocity is 450 mph on a bearing of 233.0°, and the wind's velocity is 39.0 mph from a bearing of 114.0°. Remember that "from" a direction means the wind is coming from 114.0°, so its direction of travel is 114.0° + 180° = 294.0°. Convert both velocity vectors into their component form using trigonometry. For a vector with magnitude $$v$$ and bearing $$	heta$$, the components are:

$$x = v 	imes \sin\left(\frac{\pi}{180} 	imes 	hetaight)$$

$$y = v 	imes \cos\left(\frac{\pi}{180} 	imes 	hetaight)$$

Calculate the components for the plane and the wind separately. Add the corresponding components of the plane and wind vectors to find the resultant ground velocity vector:

$$x_{result} = x_{plane} + x_{wind}$$

$$y_{result} = y_{plane} + y_{wind}$$ Calculate the magnitude of the resultant ground speed using the Pythagorean theorem:

$$	ext{Ground Speed} = \sqrt{x_{result}^2 + y_{result}^2}$$ Determine the resulting bearing of the plane by finding the angle of the resultant vector relative to north. Use the inverse tangent function:

$$	heta = \arctan\left(\frac{x_{result}}{y_{result}}ight)$$

Adjust the angle to the correct quadrant and convert it back to a bearing between 0° and 360°.

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