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A jet airliner moving initially at 369 mph to the east encounters a wind blowing at 985 mph in a direction 69◦ north of east. What is the new speed of the aircraft with respect to the ground?

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Final answer:

The new speed of the aircraft with respect to the ground is 964 mph and it is moving at an angle of 21° south of east.

Step-by-step explanation:

In order to find the new speed of the airplane with respect to the ground, we need to use vector addition.

Let's break down the velocities involved.

The initial velocity of the airplane is 369 mph to the east. We'll represent this as a vector V1 = 369 mph < 0°.

The velocity of the wind is 985 mph in a direction 69° north of east. We'll represent this as a vector Vw = 985 mph < 69°.

To find the new velocity of the airplane with respect to the ground, we can add the two vectors together using vector addition:

V_aircraft + V_wind = V_ground

V_aircraft = V_ground - V_wind

Now, let's calculate:

V_aircraft = 369 mph < 0° - 985 mph < 69°

V_aircraft = 369 mph < 0° - 985 mph < -21°

V_aircraft = (369 cos 0° - 985 cos -21°) mph < (369 sin 0° - 985 sin -21°)°

V_aircraft = 369 mph < 0° - 964 mph < -21°

Therefore, the new speed of the aircraft with respect to the ground is 964 mph and it is moving at an angle of 21° south of east.

User Kyle Trauberman
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