Answer to Question #87532 in Calculus for Kalpana

Question

A plane leaves the airport on a bearing of 45degree travelling at 400 meh. The wind is blowing at a bearing of 135 degree at a speed of 40mph.what is the actual velocity and direction of the plane?

Answer 1

$$\overrightarrow{v}=(400\cos{45^\circ}, 400\sin{45^\circ})$$

$$\overrightarrow{u}=(40\cos(-45^\circ), 40\sin(-45^\circ))$$

$$\overrightarrow{v}+\overrightarrow{u}=(400\cos{45^\circ}+40\cos(-45^\circ), 400\sin{45^\circ}+40\sin(-45^\circ))$$

$$\overrightarrow{v}+\overrightarrow{u}=(440{\sqrt{2} \over2}, 360{\sqrt{2} \over2})$$

$$|\overrightarrow{v}+\overrightarrow{u}|=\sqrt{(440{\sqrt{2} \over2})^2+(360{\sqrt{2} \over2})^2}=40\sqrt{101} \ mph\approx402 \ mph$$

$$\ bearing=\arctan{440{\sqrt{2} \over2} \over 360{\sqrt{2} \over2}}=\arctan{11 \over 9}\approx50.71^\circ$$