Question

Navigating A plane is 150 miles north of a radar station, and 30 minutes later it is 60 degree east of north at a distance of 100 miles from the radar station. Assume the plane flies on a straight line and maintains constant altitude during this 30-minute period.
a. Find the distance traveled during this 30-minute period.

Accepted Answer

First, represent the initial and final positions of the plane relative to the radar station using coordinate geometry. Place the radar station at the origin (0,0). Since the plane is initially 150 miles north, its initial coordinates are (0, 150). Next, determine the coordinates of the plane after 30 minutes. The plane is 100 miles from the radar station at a direction 60 degrees east of north. To find the coordinates, use trigonometry: the north (y) component is $$100 	imes \cos(60^\circ)$$ and the east (x) component is $$100 	imes \sin(60^\circ). So$$, the final coordinates are $$(100 \sin(60^\circ), 100 \cos(60^\circ)). $$Calculate the exact final coordinates by evaluating the sine and cosine values: $$\sin(60^\circ) = \frac{\sqrt{3}}{2}$$ and $$\cos(60^\circ) = \frac{1}{2}. $$Substitute these to get the final position as $$(100 	imes \frac{\sqrt{3}}{2}, 100 	imes \frac{1}{2}). $$Now, find the distance traveled by the plane by calculating the distance between the initial point $$(0, 150)$$ and the final point $$(100 	imes \frac{\sqrt{3}}{2}, 100 	imes \frac{1}{2}). $$Use the distance formula: $$d = \sqrt{(x_2$ - x_1$)^2$$ + ($$y_2$$ - $$y_1)^2$}. $$Substitute the coordinates into the distance formula: $$d = \sqrt{\left(100 	imes \frac{\sqrt{3}}{2} - 0\$$right)^2$$ + \left(100 	imes \frac{1}{2} - 150\$$right)^2$$}. $$Simplify inside the square root to express the distance traveled.

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Key Concepts

Vector Representation and Components

Law of Cosines

Distance and Displacement in Plane Geometry