Question

A plane is headed due south with an airspeed of 192 mph. A wind from a direction of 78.0° is blowing at 23.0 mph. Find the ground speed and resulting bearing of the plane.

Accepted Answer

Identify the vectors involved: The plane's velocity vector is due south with a magnitude of 192 mph, and the wind's velocity vector has a magnitude of 23 mph coming from 78.0°. Note that "from 78.0°" means the wind is blowing towards 78.0° + 180° = 258.0° (since wind direction is given as the direction it comes from). Express both velocity vectors in component form using the standard coordinate system where 0° is east and angles increase counterclockwise. For the plane headed due south, its velocity vector components are $$V_p = (0, -192)$$ mph. For the wind, convert the direction 258.0° to components: $$V_w = (23 \cos(258^\circ), 23 \sin(258^\circ))$$ mph. Add the two vectors component-wise to find the resultant ground velocity vector: $$V_g = V_p + V_w = (0 + 23 \cos(258^\circ), -192 + 23 \sin(258^\circ)). $$Calculate the magnitude of the ground velocity vector (ground speed) using the Pythagorean theorem: $$|V_g| = \sqrt{(V_{gx})^2$ + (V_{gy})^2$}$$, where $$V_{gx}$$ and $$V_{gy}$$ are the x and y components of $$V_g. $$Determine the bearing (direction) of the ground velocity vector by finding the angle $$	heta it $$makes with the north direction. Use the inverse tangent function: $$	heta = \arctan\left(\frac{|V_{gx}|}{|V_{gy}|}ight)$$, then adjust the angle based on the quadrant to express the bearing as an angle clockwise from north.

A plane is headed due south with an airspeed of 192 mph. A wind from a direction of 78.0° is blowing at 23.0 mph. Find the ground speed and resulting bearing of the plane.

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