A jet airliner moving at 470 mph due east moves into a region where the wind is blowing at 110 mph in a direction 30.0° north of east.What is the new velocity and direction of the aircraft?
A)568 mph,5.56° N of E
B)573 mph,5.98° N of E
C)520 mph,9.06° N of E
D)568 mph,10.4° N of E
E)448 mph,9.45° N of E

Question

Quizplus · Accepted Answer

The new velocity and direction can be found using vector addition. The eastward component of the wind's velocity is $110 \cos(30^\circ)$ mph and the northward component is $110 \sin(30^\circ)$ mph. Adding these to the plane's velocity gives a total eastward velocity of $470 + 110 \cos(30^\circ)$ mph and a northward velocity of $110 \sin(30^\circ)$ mph. The magnitude of the resultant velocity is $\sqrt{(470 + 110 \cos(30^\circ))^2 + (110 \sin(30^\circ))^2}$ mph, which calculates to approximately 568 mph. The direction, $\theta$, can be found using the tangent function: $\theta = \tan^{-1}\left(\frac{110 \sin(30^\circ)}{470 + 110 \cos(30^\circ)}\right)$, which gives about 5.56° north of east.

A Jet Airliner Moving at 470 Mph Due East Moves