Final answer:
The speed of the plane with respect to the ground is 95.39 m/s, its heading is 20.93° east of due north, and in one hour it will travel 324 km eastward.
Step-by-step explanation:
To solve this problem, we need to use vector addition to combine the velocity of the plane relative to the air with the velocity of the air relative to the ground. We're given that the plane's velocity relative to the air is 110.0 m/s due east and the air's velocity is 40.0 m/s at an angle of 30° west of due north.
1) To find the speed of the plane with respect to the ground, we decompose the air's velocity into two components: northward and westward. The northward component is 40.0 m/s * cos(30°) and the westward component is 40.0 m/s * sin(30°). We then add the eastward velocity of the plane to the westward component of the wind's velocity (keeping in mind that eastward is positive, and westward is negative) to get the eastward component of the ground velocity.
Northward component of air's velocity = 40.0 * cos(30°) = 34.64 m/s northward
Westward component of air's velocity = 40.0 * sin(30°) = 20.0 m/s westward
Eastward velocity of the plane relative to the ground = 110.0 m/s - 20.0 m/s = 90.0 m/s eastward
The resultant ground velocity is thus the vector sum of the northward and eastward components. Using the Pythagorean theorem, the speed (magnitude of velocity) of the plane with respect to the ground is:
Plane's speed relative to ground = √(34.64² + 90.0²) = 95.39 m/s
2) To determine the heading of the plane with respect to the ground, we calculate the inverse tangent of the ratio of the northward component to the eastward component.
Heading angle = tan⁻¹(34.64/90.0) = 20.93° east of due north
3) In one hour, the distance the plane travels eastward will be the eastward component of the ground velocity multiplied by the total time, considering that 1 hour is equal to 3600 seconds.
Distance Eastward = 90.0 m/s * 3600 s = 324,000 meters or 324 km eastward