Use the given flight data to compute the ground speed, the drift angle, and the course (round your answers to two decimal places). The heading and air speed are $45^{\circ}$ and $275 \mathrm{mph}$ , respectively, the wind is $50 \mathrm{mph}$ from $135^{\circ}$ .

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Q. Use the given flight data to compute the ground speed, the drift angle, and the course (round your answers to two decimal places). The heading and air speed are

45^{\circ}

and

275 \mathrm{mph}

, respectively, the wind is

50 \mathrm{mph}

from

135^{\circ}

Vector Addition Explanation: To solve this problem, we will use vector addition to combine the velocity vector of the aircraft with the wind vector to find the resulting ground speed and direction. The aircraft's velocity vector is $275 \, \text{mph}$ at a heading of $45$ degrees. The wind's velocity vector is $50 \, \text{mph}$ from $135$ degrees, which means it is blowing towards the southwest. We will convert these headings into standard mathematical angles measured counterclockwise from the positive x-axis. $\newline$ Aircraft heading angle: $45$ degrees (no conversion needed as it is already in the standard mathematical direction). $\newline$ Wind angle: $135$ degrees is equivalent to $180 - 135 = 45$ degrees measured clockwise from the negative x-axis.
Convert Headings: Next, we will break down the aircraft's velocity and the wind's velocity into their respective components along the x (east-west) and y (north-south) axes. $\newline$ Aircraft velocity components: $\newline$ $V_x (\text{aircraft}) = 275 \, \text{mph} \cdot \cos(45^\circ)$ $\newline$ $V_y (\text{aircraft}) = 275 \, \text{mph} \cdot \sin(45^\circ)$ $\newline$ Wind velocity components: $\newline$ $V_x (\text{wind}) = 50 \, \text{mph} \cdot \cos(45^\circ)$ (since the wind is from $135^\circ$ , this component will be negative) $\newline$ $V_y (\text{wind}) = 50 \, \text{mph} \cdot \sin(45^\circ)$ (since the wind is from $135^\circ$ , this component will also be negative)
Calculate Components: Now we will calculate the components. $\newline$ $V_x$ (aircraft) $= 275$ mph $\times \cos(45$ degrees) $= 275$ mph $\times (\sqrt{2}/2) \approx 194.66$ mph $\newline$ $V_y$ (aircraft) $= 275$ mph $\times \sin(45$ degrees) $= 275$ mph $\times (\sqrt{2}/2) \approx 194.66$ mph $\newline$ $V_x$ (wind) $= 275$ $1$ mph $\times \cos(45$ degrees) $= 275$ $1$ mph $= 275$ $4$ mph $\newline$ $V_y$ (wind) $= 275$ $1$ mph $\times \sin(45$ degrees) $= 275$ $1$ mph $= 275$ $4$ mph
Add Velocity Components: We will now add the components of the aircraft's velocity and the wind's velocity to find the ground speed components. $\newline$ $V_x (\text{ground}) = V_x (\text{aircraft}) + V_x (\text{wind}) = 194.66 \, \text{mph} - 35.36 \, \text{mph} \approx 159.30 \, \text{mph}$ $\newline$ $V_y (\text{ground}) = V_y (\text{aircraft}) + V_y (\text{wind}) = 194.66 \, \text{mph} - 35.36 \, \text{mph} \approx 159.30 \, \text{mph}$
Calculate Ground Speed: The ground speed magnitude can be found by calculating the resultant vector's magnitude using the Pythagorean theorem. $\newline$ Ground speed = $\sqrt{(V_{x} \text{(ground)})^2 + (V_{y} \text{(ground)})^2}$ $\newline$ Ground speed = $\sqrt{(159.30 \, \text{mph})^2 + (159.30 \, \text{mph})^2}$
Find Drift Angle: Calculating the ground speed magnitude: $\newline$ Ground speed = $\sqrt{(159.30 \, \text{mph})^2 + (159.30 \, \text{mph})^2} = \sqrt{2 \times (159.30 \, \text{mph})^2} = 159.30 \, \text{mph} \times \sqrt{2} \approx 225.35 \, \text{mph}$
Calculate Course: To find the drift angle, we need to calculate the angle of the ground speed vector relative to the aircraft's heading. This can be done using the $\text{arctan}$ function to find the angle between the resultant ground speed vector and the positive x-axis, and then adjusting for the aircraft's initial heading. $\text{Drift angle} = \text{arctan}(\frac{V_y \text{(ground)}}{V_x \text{(ground)}}) - \text{aircraft heading}$ $\text{Drift angle} = \text{arctan}(\frac{159.30 \text{ mph}}{159.30 \text{ mph}}) - 45 \text{ degrees}$
Calculate Course: To find the drift angle, we need to calculate the angle of the ground speed vector relative to the aircraft's heading. This can be done using the $\text{arctan}$ function to find the angle between the resultant ground speed vector and the positive x-axis, and then adjusting for the aircraft's initial heading. $\text{Drift angle} = \text{arctan}(\frac{V_y \text{(ground)}}{V_x \text{(ground)}}) - \text{aircraft heading}$ $\text{Drift angle} = \text{arctan}(\frac{159.30 \text{ mph}}{159.30 \text{ mph}}) - 45 \text{ degrees}$ Calculating the drift angle: $\text{Drift angle} = \text{arctan}(\frac{159.30 \text{ mph}}{159.30 \text{ mph}}) - 45 \text{ degrees} = \text{arctan}(1) - 45 \text{ degrees} = 45 \text{ degrees} - 45 \text{ degrees} = 0 \text{ degrees}$ The drift angle is $0 \text{ degrees}$ , which means there is no drift, and the aircraft's course is the same as its heading.
Calculate Course: To find the drift angle, we need to calculate the angle of the ground speed vector relative to the aircraft's heading. This can be done using the $\text{arctan}$ function to find the angle between the resultant ground speed vector and the positive x-axis, and then adjusting for the aircraft's initial heading. $\text{Drift angle} = \text{arctan}(\frac{V_y \text{(ground)}}{V_x \text{(ground)}}) - \text{aircraft heading}$ $\text{Drift angle} = \text{arctan}(\frac{159.30 \text{ mph}}{159.30 \text{ mph}}) - 45 \text{ degrees}$ Calculating the drift angle: $\text{Drift angle} = \text{arctan}(\frac{159.30 \text{ mph}}{159.30 \text{ mph}}) - 45 \text{ degrees} = \text{arctan}(1) - 45 \text{ degrees} = 45 \text{ degrees} - 45 \text{ degrees} = 0 \text{ degrees}$ The drift angle is $0 \text{ degrees}$ , which means there is no drift, and the aircraft's course is the same as its heading.Finally, we will calculate the course. Since the drift angle is $0 \text{ degrees}$ , the course is the same as the heading. $\text{Course} = \text{aircraft heading} + \text{drift angle}$ $\text{Course} = 45 \text{ degrees} + 0 \text{ degrees} = 45 \text{ degrees}$