Question

A plane has an airspeed of 520 mph. The pilot wishes to fly on a bearing of 310°. A wind of 37 mph is blowing from a bearing of 212°. In what direction should the pilot fly, and what will be her ground speed?

Accepted Answer

Identify the vectors involved: the plane's airspeed vector (magnitude 520 mph, direction to be found), the wind vector (magnitude 37 mph, coming from bearing 212°), and the resultant ground speed vector (desired bearing 310°). Convert the wind's bearing to a direction the wind is blowing towards by adding 180° to the wind's bearing (since wind direction is given as where it comes from). Calculate the wind vector components using $$ V_{wind_x} = 37 	imes \cos\left(\frac{(212 + 180) 	imes \pi}{180}ight) $$ and $$ V_{wind_y} = 37 	imes \sin\left(\frac{(212 + 180) 	imes \pi}{180}ight) . $$Express the desired ground speed vector components using the bearing 310°: $$ V_{ground_x} = V_{ground} 	imes \cos\left(\frac{310 	imes \pi}{180}ight) $$ and $$ V_{ground_y} = V_{ground} 	imes \sin\left(\frac{310 	imes \pi}{180}ight) $$, where $$ V_{ground}  is $$the unknown ground speed magnitude. Set up the vector equation for the plane's airspeed vector $$ \vec{V}_{plane} $$ such that $$ \vec{V}_{plane} + \vec{V}_{wind} = \vec{V}_{ground} . $$Using components, this becomes two equations: $$ V_{plane_x} + V_{wind_x} = V_{ground_x} $$ and $$ V_{plane_y} + V_{wind_y} = V_{ground_y} . $$Use the known magnitude of the plane's airspeed (520 mph) and the two component equations to solve for the plane's heading angle (direction to fly) and the ground speed magnitude $$ V_{ground} . $$This involves expressing $$ V_{plane_x} = 520 	imes \cos	heta $$ and $$ V_{plane_y} = 520 	imes \sin	heta $$, then solving the system for $$ 	heta $$ and $$ V_{ground} .$$

A plane has an airspeed of 520 mph. The pilot wishes to fly on a bearing of 310°. A wind of 37 mph is blowing from a bearing of 212°. In what direction should the pilot fly, and what will be her ground speed?

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