91Ó°ÊÓ

In the context of the wind velocity problem, vector displacement is a key concept used to understand how the location of the airplane changes due to both its motion and external forces like wind. A vector displacement describes both the distance and direction of an object from a point of origin. It has both horizontal and vertical components that can be expressed using unit vectors.
For example, in this exercise, the vector displacement of the airplane includes a westward component of 120 km and a southward component of 20 km. Together, the vector displacement can be expressed as the sum of these individual components: \(\vec{d}_{actual} = 120 \hat{i} + 20 \hat{j}\). Here, \(\hat{i}\) represents the westward direction, and \(\hat{j}\) represents the southward direction. By understanding vector displacement, one can analyze how different forces act on the airplane during its flight.

Airspeed calculation is crucial for determining how the airplane moves relative to the surrounding air. It is the speed at which the airplane would fly in the absence of wind effects.
To calculate the intended displacement under just its own power, you multiply the airspeed by the time the airplane is in motion. In our problem, the airplane was flying due west with an airspeed of 220 km/h for 0.5 hours. Therefore, the calculated intended displacement was: \(\vec{d}_{intended} = 220 \times 0.5 = 110 \hat{i}\) \text{ km}.
Understanding airspeed calculations allows pilots to plan courses in a way that can efficiently counteract wind effects, ensuring the aircraft reaches its intended destination on time.

Determining the angle of direction is essential for understanding where the wind or airplane is heading relative to different axes. In this case, it involves trigonometric calculations to find the angle between the wind velocity and westward direction.
Using the components of the wind velocity vector \(\vec{v}_{wind} = 20 \hat{i} + 40 \hat{j}\), the angle of direction can be calculated using the inverse tangent function: \(\theta = \tan^{-1}\left(\frac{40}{20}\right) = \tan^{-1}(2) \approx 63.43^\circ\). This angle indicates the direction of the wind as \(63.43^\circ\) south of west.
Understanding angles in vector problems enables better navigation and course corrections by accounting for crosswinds and other directional forces affecting an airplane's path.

The Pythagorean theorem is a fundamental principle used to calculate the magnitude of a vector by combining its components. It states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
For the wind velocity vector, which has components in both the \(\hat{i}\) and \(\hat{j}\) directions, the Pythagorean theorem helps find the resultant magnitude. Thus, the magnitude of the wind velocity vector is calculated as: \(v_{wind} = \sqrt{20^2 + 40^2} = \sqrt{400 + 1600} = \sqrt{2000} \approx 44.72 \, \text{km/h}\).
This calculation showcases the total influence of wind on the airplane's path, represented as a single speed value.

Trigonometry in physics is pivotal in resolving vectors and analyzing forces and motions. Trigonometric functions like sine, cosine, and tangent help relate angles to the sides of triangles, which are analogous to vector components in many physics problems.
In our airplane navigation scenario, trigonometry is employed to adjust the pilot's course due to wind. For instance, if the wind velocity is 40 km/h south, the pilot uses the sine function \(\sin(\theta) = \frac{40}{220}\) to find the angle \(\theta\) required to maintain a course due west against the wind's force, resulting in \(\theta = \sin^{-1}\left(\frac{40}{220}\right) \approx 10.47^\circ\). This means the pilot must steer slightly north of west to compensate for the southerly wind.
Trigonometry not only aids in calculating angles but also assists in understanding and overcoming the various forces that objects encounter, making it indispensable in physics.