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In physics, when dealing with motions like that of an airplane, it's crucial to break down velocity into components. This process helps us understand how the airplane is moving in different directions. Typically, we look at motion in the north-south and east-west directions, using trigonometry to assist us.

For an airplane flying at a speed of 550 mph towards a bearing of \(43^\circ\) east of north, we split this velocity into northward and eastward components:

• Northward (\(v_{a_n}\)): \(v_{a_n} = 550 \times \cos(43^\circ)\)
• Eastward (\(v_{a_e}\)): \(v_{a_e} = 550 \times \sin(43^\circ)\)

Calculating these, we find the plane moves approximately 402.59 mph north and 375.24 mph east. This decomposition is vital as it simplifies combined velocity effects from both the airplane's desired path and external factors like wind.

Wind can significantly alter an object’s trajectory, and understanding its impact is essential in navigation and aviation. In scenarios like the given exercise, wind blowing from the southwest affects the airplane's original flight path.

The wind speed here is 25 mph, moving \(15^\circ\) east of north from the southwest. You'll need to calculate these components to see how it modifies the airplane's course:

• Northward wind component (\(v_{w_n}\)): \(v_{w_n} = 25 \times \cos(15^\circ)\)
• Eastward wind component (\(v_{w_e}\)): \(v_{w_e} = 25 \times \sin(15^\circ)\)

After computation, the northward component is approximately 24.15 mph, and the eastward is about 6.47 mph. These values indicate how much the wind alters the plane's north and east velocities.

To find the airplane’s final speed, or ground speed, you combine the velocity components of the airplane and the wind. This step involves vector addition.

First, calculate the ground velocity components:

• Northward: \(v_{g_n} = v_{a_n} + ( -v_{w_n} ) = 378.44\)
• Eastward: \(v_{g_e} = v_{a_e} + (-v_{w_e}) = 368.77\)

The negative sign signifies that wind velocity opposes the airplane's movement in both directions. Finally, to discover the ground speed, apply the Pythagorean theorem: \[\text{Ground speed} = \sqrt{378.44^2 + 368.77^2} \approx 528.06 \text{ mph}\] This value tells us how fast the airplane is moving over the ground, considering the original velocity and wind influence.

Determining the bearing or direction involves calculating the angle of the final velocity vector with respect to north. Such measurements help pilots adjust their navigation to stay on course. Using trigonometry, more specifically, the tangent function, simplifies this task.

To find the angle \(\theta\), utilize:

• \(\theta = \tan^{-1}\left(\frac{v_{g_e}}{v_{g_n}}\right)\)

Performing the calculation with the given components, \(v_{g_e} = 368.77\) and \(v_{g_n} = 378.44\), results in an angle: \[\theta \approx 44.01^\circ\] This angle represents the airplane's new bearing. Thus, due to wind effects, the aircraft is now heading \(44.01^\circ\) east of north.