To solve the problem of finding the ground speed and actual direction of the airplane, we will use vector addition. The plane's velocity and the wind's velocity are combined to find the resultant vector, which represents the ground velocity.

Step 1: Represent the velocities as vectors

The velocity vectors will be represented in terms of their components in the north-east coordinate system:

• Bearing $300^\circ$ means $360^\circ - 300^\circ = 60^\circ$ west of north.
• Bearing $200^\circ$ means $200^\circ - 180^\circ = 20^\circ$ south of west.

Plane's velocity vector:

The magnitude is $560 \, \text{mph}$, and the bearing is $300^\circ$: $V_{\text{plane}} = (560 \cos(300^\circ), 560 \sin(300^\circ)).$ Using trigonometric values: $\cos(300^\circ) = \cos(-60^\circ) = 0.5, \quad \sin(300^\circ) = \sin(-60^\circ) = -\frac{\sqrt{3}}{2}.$ $V_{\text{plane}} = (560 \times 0.5, 560 \times -\frac{\sqrt{3}}{2}) = (280, -484.12).$

Wind's velocity vector:

The magnitude is $20 \, \text{mph}$, and the bearing is $200^\circ$: $V_{\text{wind}} = (20 \cos(200^\circ), 20 \sin(200^\circ)).$ Using trigonometric values: $\cos(200^\circ) = \cos(180^\circ + 20^\circ) = -\cos(20^\circ), \quad \sin(200^\circ) = \sin(180^\circ + 20^\circ) = -\sin(20^\circ).$ Approximating the trigonometric values: $\cos(20^\circ) \approx 0.9397, \quad \sin(20^\circ) \approx 0.3420.$ $V_{\text{wind}} = (20 \times -0.9397, 20 \times -0.3420) = (-18.79, -6.84).$

Step 2: Add the velocity vectors

The ground velocity vector $V_{\text{ground}}$ is the sum of the plane's and wind's velocity vectors: $V_{\text{ground}} = V_{\text{plane}} + V_{\text{wind}}.$ $V_{\text{ground}} = (280 + (-18.79), -484.12 + (-6.84)).$ $V_{\text{ground}} = (261.21, -490.96).$

Step 3: Calculate the ground speed

The ground speed is the magnitude of the resultant vector: $\text{Ground speed} = \sqrt{(261.21)^2 + (-490.96)^2}.$ $\text{Ground speed} = \sqrt{68191.06 + 241043.32} = \sqrt{309234.38}.$ $\text{Ground speed} \approx 556.06 \, \text{mph}.$

Step 4: Find the actual direction

The actual direction is the angle of the resultant vector from the north, given by: $\theta = \tan^{-1}\left(\frac{\text{east component}}{\text{north component}}\right).$ $\theta = \tan^{-1}\left(\frac{261.21}{-490.96}\right).$ $\theta = \tan^{-1}(-0.532).$ $\theta \approx -27.99^\circ.$

Since the angle is measured clockwise from north: $\text{Actual direction} = 360^\circ - 27.99^\circ = 332.01^\circ.$

Final Answer:

• Ground speed: $556.06 \, \text{mph}$.
• Actual direction: $332.01^\circ$.

Would you like further clarification on any step?

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5 Related Questions:

1. How do you find the bearing if the wind direction changes?
2. What happens to the ground speed if the wind speed increases?
3. How would the calculations differ if the wind had a bearing of $0^\circ$ (north)?
4. Why do we use vector addition to solve this type of problem?
5. How would you calculate the airspeed if only the ground speed and wind speed were known?

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Tip:

Always break velocity into components for easier calculation when working with bearings and directions!