You are on an airplane traveling 30° south of due west at 130 m/s with respect to the air. The air is moving with a speed 30 m/s with respect to the ground due north.
One question..
2)
What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due easth).
I'm having some trouble figuring out this question. If anyone could help me out it would be greatly appreciated. Thanks
speed north = 30 - 130 sin 30 = -35
speed west = 130 cos 30 = 113
tan angle south of west = 35/113
so angle south of west = 17.2 deg
so 270 - 17.2 = 253 on the compass rose
Well, isn't this a bit of a headscratcher! Let's see if I can help you out with a dash of humor.
So, you're on an airplane that's heading 30° south of due west, and the air beneath you is feeling a little speedy as it moves 30 m/s due north. Now, imagine the airplane wants to have a chat with the ground and figure out its heading. How polite!
To find the heading of the plane with respect to the ground, we can think of it as a little game of "Wind vs. Plane." The plane is going 30° south of due west, while the air is hustling along at 30 m/s north. It's almost like a dance-off, but without all the fancy twirls.
Now, let's imagine this airplane deciding to boogie its way through the air. It's going west but getting pushed north by the air. And when it comes down to the nitty-gritty, that combination of directions gives us the answer we're looking for – the ground direction the plane is heading towards.
If we crunch the numbers, the angle between the plane's heading and the ground will be the arctan of the ground speed divided by the airspeed. So, in this case, that's the arctan of 30 m/s divided by 130 m/s (the ground speed of the plane), which will give us the angle.
Just remember, my friend, to convert your answer to degrees so it fits our lovely 0° representing due north and 90° representing due east scale. I hope this helps, and I applaud your determination to solve this puzzling question! Safe travels!
To determine the heading of the plane with respect to the ground, we need to find the resultant velocity vector by combining the velocity of the plane with respect to the air and the velocity of the air with respect to the ground.
Given:
- Velocity of the plane with respect to the air (v_airplane): 130 m/s at a bearing of 30° south of due west.
- Velocity of the air with respect to the ground (v_air): 30 m/s due north.
We can break down the velocity vectors into their components:
- Velocity of the plane with respect to the air (v_airplane):
- Horizontal component (v_airplane_x): v_airplane * cos(30°)
- Vertical component (v_airplane_y): -v_airplane * sin(30°)
- Velocity of the air with respect to the ground (v_air):
- Horizontal component (v_air_x): 0 (as it is due north)
- Vertical component (v_air_y): v_air
To find the resultant velocity vector, we add the horizontal and vertical components:
- Resultant horizontal velocity (v_resultant_x): v_airplane_x + v_air_x
- Resultant vertical velocity (v_resultant_y): v_airplane_y + v_air_y
Finally, we can calculate the magnitude and direction (heading) of the resultant velocity:
- Magnitude (v_resultant): sqrt(v_resultant_x^2 + v_resultant_y^2)
- Direction (θ): atan2(v_resultant_y, v_resultant_x), with the result in degrees
Let's calculate it step by step:
Step 1: Calculate the components of the velocity:
v_airplane_x = 130 m/s * cos(30°)
v_airplane_y = -130 m/s * sin(30°)
v_air_x = 0
v_air_y = 30 m/s
Step 2: Calculate the resultant components:
v_resultant_x = v_airplane_x + v_air_x
v_resultant_y = v_airplane_y + v_air_y
Step 3: Calculate the magnitude and direction:
v_resultant = sqrt(v_resultant_x^2 + v_resultant_y^2)
θ = atan2(v_resultant_y, v_resultant_x)
Step 4: Convert the direction to the range of 0° to 360°:
if θ < 0, θ += 360°
Let's substitute the values and calculate the result:
v_airplane_x = 130 m/s * cos(30°) ≈ 112.39 m/s
v_airplane_y = -130 m/s * sin(30°) ≈ -65 m/s
v_air_x = 0 m/s
v_air_y = 30 m/s
v_resultant_x = 112.39 m/s + 0 m/s ≈ 112.39 m/s
v_resultant_y = -65 m/s + 30 m/s ≈ -35 m/s
v_resultant = sqrt((112.39 m/s)^2 + (-35 m/s)^2) ≈ 118.14 m/s
θ = atan2(-35 m/s, 112.39 m/s) ≈ -17.4°
Since the direction is negative, we add 360° to obtain the final result:
θ_final = -17.4° + 360° ≈ 342.6°
Therefore, the heading of the plane with respect to the ground is approximately 342.6°.
To find the heading of the plane with respect to the ground, we can use vector addition.
First, let's break down the velocity of the plane into its horizontal and vertical components. Since the plane is traveling 30° south of due west, we can split the velocity vector into two components: one in the westward direction and one in the southward direction.
The westward component of the plane's velocity can be found by multiplying the magnitude of the plane's velocity (130 m/s) by the cosine of the angle it makes with the westward direction.
So, the westward component of the plane's velocity is:
130 m/s * cos(30°) = 112.45 m/s (approximately)
The southward component of the plane's velocity can be found by multiplying the magnitude of the plane's velocity (130 m/s) by the sine of the angle it makes with the westward direction.
So, the southward component of the plane's velocity is:
130 m/s * sin(30°) = 65 m/s (approximately)
Now, let's consider the velocity of the air. The air is moving with a speed of 30 m/s due north.
To find the total velocity of the plane with respect to the ground, we need to add the velocity of the plane with respect to the air to the velocity of the air.
Since the air is moving due north and the plane is moving towards the southwest, their velocities are perpendicular to each other. So, we can use the Pythagorean theorem to find the magnitude of their combined velocity:
Magnitude of combined velocity = √(westward component^2 + (air velocity)^2)
Substituting the values:
Magnitude of combined velocity = √(112.45^2 + 30^2) = √(12623 + 900) = √13523 ≈ 116.26 m/s
The orientation of the combined velocity can be found using trigonometry.
The angle of the combined velocity relative to the north direction can be found by taking the inverse tangent of the southward component divided by the westward component:
Angle = arctan(southward component / westward component)
= arctan(65 m/s / 112.45 m/s)
≈ 29.7°
Since the angle represents the direction clockwise with respect to due north, we need to subtract it from 90° to get the heading of the plane with respect to the ground.
Heading = 90° - 29.7°
= 60.3° (approximately)
Therefore, the heading of the plane with respect to the ground is approximately 60.3° (counterclockwise from due east).