We can use vector addition to solve this problem. Let's first draw a diagram to visualize the situation:
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| 413 mph, due West
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| Wind: 15 mph
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The vector representing the airplane's velocity is 413 mph due West, which means it has no component in the North-South direction. The vector representing the wind is 15 mph in a direction 45 degrees South of West, which means it has components in both the West and South directions.
To find the resulting velocity of the airplane relative to the ground, we need to add these two vectors. We can break the wind vector into its West and South components using trigonometry:
West component = 15 mph * cos(45) = 10.61 mph
South component = 15 mph * sin(45) = 10.61 mph
Now we can add the two vectors:
Resultant West component = 413 mph + 10.61 mph = 423.61 mph
Resultant South component = 0 mph - 10.61 mph = -10.61 mph
The negative sign on the South component indicates that the airplane is traveling South relative to its starting point. To find the bearing of the airplane in degrees relative to the positive x-axis, we need to find the angle between the resultant velocity vector and the positive x-axis.
We can use trigonometry to find this angle:
tan(theta) = opposite/adjacent
theta = atan(opposite/adjacent)
In this case, the opposite side is the South component (-10.61 mph) and the adjacent side is the West component (423.61 mph). So we have:
theta = atan(-10.61/423.61) = -1.43 degrees
Note that the negative sign indicates that the angle is measured clockwise from the positive x-axis. To get the angle measured counterclockwise, we can add 360 degrees:
theta = -1.43 + 360 = 358.57 degrees
So the bearing of the airplane in degrees relative to the positive x-axis is approximately 358.57 degrees (rounded to two decimal places).
Hope this helps,
- Jeron :)