a twin-engine cessna is heading on bearing of 35 degrees with an air speed of 180 mph. if the wind is out with a bearing of 90 degrees at 40 mph, then what is the bearing of this course? what is the ground seed of the airplane?
vehicles and winds travel on a heading, not a bearing. Sadly, this distinction is ignored or unknown to many textbook writers and teachers.
But on to the math.
the Cessna in still air travels be at
(180 sin35°,180 cos35°) = (103.24,147.45)
The wind blows it an additional (-90,0)
Thus, its final velocity is (13.24,147.45)
Thus, its ground speed is √(13.24^2 + 147.45^2) = 148.04
and its heading is 5.2°
To find the bearing of the course, we need to add the wind vector to the airplane's heading vector. This can be done using vector addition. Here are the steps:
1. Convert the given bearings into vector form. The airplane heading vector has a magnitude of 180 mph and a direction of 35 degrees. The wind vector has a magnitude of 40 mph and a direction of 90 degrees.
2. Split the airplane heading vector into its north and east components using trigonometry. The north component can be calculated by multiplying the magnitude (180 mph) by the cosine of the bearing (35 degrees), and the east component can be calculated by multiplying the magnitude by the sine of the bearing.
3. Split the wind vector in the same way. The north component can be calculated by multiplying the magnitude (40 mph) by the cosine of the bearing (90 degrees), and the east component is zero since the wind is blowing directly from the east.
4. Add the north components and east components separately. The resulting vectors will give us the new heading vector.
5. Convert the new heading vector from component form to magnitude and bearing form using trigonometry. The magnitude can be calculated using the Pythagorean theorem, and the bearing can be calculated using the arctan function.
6. The bearing of the course is the bearing of the new heading vector. The ground speed of the airplane is the magnitude of the new heading vector.
Applying these steps:
1. The airplane heading vector is (180 mph, 35 degrees).
The wind vector is (40 mph, 90 degrees).
2. The north component of the airplane heading vector is given by: 180 mph * cos(35 degrees) = 146.32 mph.
The east component is given by: 180 mph * sin(35 degrees) = 101.21 mph.
3. The north component of the wind vector is given by: 40 mph * cos(90 degrees) = 0 mph.
The east component is zero since the wind is blowing directly from the east.
4. Adding the north components: 146.32 mph + 0 mph = 146.32 mph.
Adding the east components: 101.21 mph + 0 mph = 101.21 mph.
5. The new heading vector is (146.32 mph, 101.21 mph).
Its magnitude is given by: √(146.32^2 + 101.21^2) ≈ 177.61 mph.
Its bearing is given by: arctan(101.21 mph / 146.32 mph) ≈ 35.96 degrees.
6. Therefore, the bearing of the course is approximately 35.96 degrees.
The ground speed of the airplane is approximately 177.61 mph.