A jet airliner moving at 600 mph due east moves into a region where the wind is blowing at 120 mph in a direction 30.0° north of east.What is the new speed and direction of the aircraft?
A) 706 mph,5.67° N of E
B) 620 mph,5.67° N of E
C) 706 mph,4.87° N of E
D) 588 mph,4.87° N of E

Question

Quizplus · Accepted Answer

To solve the problem, we need to use vector addition. The velocity of the plane relative to the ground can be expressed as the sum of its velocity in still air (600 mph due east) and the velocity of the wind (120 mph 30° north of east).

Using trigonometry, we can resolve the wind velocity into its components:
wind north = 120 sin 30° = 60 mph
wind east = 120 cos 30° = 103.92 mph

We can now add the velocities vectorially. Let's call the resulting velocity V.
V north = plane north + wind north = 0 + 60 = 60 mph north
V east = plane east + wind east = 600 + 103.92 = 703.92 mph east

The magnitude of V is the square root of (V north)^2 + (V east)^2 = sqrt(60^2 + 703.92^2) = 706 mph.

The direction of V can be found as the inverse tangent of (V north / V east), which gives us 4.87° north of east.

Therefore, the new speed and direction of the aircraft are 706 mph, 4.87° north of east. Answer choice C is the correct answer.

A Jet Airliner Moving at 600 Mph Due East Moves