I'm having trouble figuring this out. An airplane has an airspeed of 160 mph. It is to make a flight in a direction of 80° while there's a 20 mph wind from 310°. What will the airplane's actual heading be?
I'd appreciate any help especially on how you determined what angles to use. Thank you.
just convert everything to x-y values, add them up, and change back to headings. If his heading is θ, then we need
160sinθ - 20sin310° = v*sin80°
160cosθ - 20cos310° = v*cos80°
solve that and you get
θ = 74.5°
v = 172.1
makes sense, since the wind is blowing to the SE, adding to his speed.
To determine the airplane's actual heading, we need to find the direction of the airplane's resultant velocity vector.
First, let's draw a diagram to understand the situation better.
20 mph (310°) ^
| |
| |
| |
____|____ | 160 mph
| | |
| | |
| A | |
|plane |------------> V (Resultant Velocity)
| | 80° |
In this diagram, the vector A represents the wind's velocity of 20 mph at an angle of 310°, and V represents the resultant velocity vector of the airplane.
To determine the resultant velocity vector, we can use vector addition. We can break down the airspeed vector into its x and y components.
Given that the airspeed of the airplane is 160 mph and that it is making a flight in a direction of 80°, we can calculate the x and y components as:
Vx = 160 mph * cos(80°)
Vy = 160 mph * sin(80°)
Next, we need to consider the wind's velocity vector, A. The wind is blowing at 20 mph at an angle of 310°. We can calculate the x and y components of this vector as:
Ax = 20 mph * cos(310°)
Ay = 20 mph * sin(310°)
To find the resultant velocity vector, we add the components of the airspeed vector and the wind vector:
Vx' = Vx + Ax
Vy' = Vy + Ay
Now, we can find the magnitude and direction of the resultant velocity vector using the Pythagorean theorem and inverse trigonometric functions:
Magnitude (V') = √(Vx'² + Vy'²)
Direction (θ) = arctan(Vy' / Vx')
The direction of the airplane's actual heading will be the angle θ.
By substituting the calculated values, we can find the resultant velocity's magnitude and direction:
Magnitude (V') = √((160 mph * cos(80°) + 20 mph * cos(310°))² + (160 mph * sin(80°) + 20 mph * sin(310°))²)
Direction (θ) = arctan((160 mph * sin(80°) + 20 mph * sin(310°)) / (160 mph * cos(80°) + 20 mph * cos(310°)))
Calculating these values will give you the airplane's actual heading.