An airplane has a velocity of 580 km/h relative to the moving air. At the same time, a wind blows northward with a speed of 92 km/h relative to earth. The airplane is moving in an easterly direction relative to earth. Find the speed of the airplane to relative to earth (answer in terms of degrees south of east).

To find the speed of the airplane relative to the Earth, we need to take into account the velocity of the airplane and the velocity of the wind.

Let's break it down into components. The velocity of the airplane relative to Earth can be divided into two components: one in the east-west direction and one in the north-south direction. Similarly, we can break down the velocity of the wind into its components.

Given:
Velocity of the airplane relative to the moving air = 580 km/h (which is also the east-west component of the airplane's velocity)
Velocity of the wind relative to Earth = 92 km/h (which is the northward component of the wind's velocity)

Since the airplane is moving eastward (in the positive east-west direction), the east-west component of its velocity relative to Earth remains the same as the velocity relative to the moving air: 580 km/h.

To find the north-south component of the airplane's velocity relative to the Earth, we need to subtract the northward component of the velocity of the wind from the north-south component of the airplane's velocity relative to the moving air.

Let's denote the north-south component of the airplane's velocity relative to the moving air as Vy_airplane and the north-south component of the wind's velocity as Vy_wind.

Given:
Vy_wind = 0 km/h (since the wind is blowing strictly northward)

Therefore, Vy_airplane = Vy_airplane - Vy_wind = Vy_airplane - 0 = Vy_airplane.

So, the north-south component of the airplane's velocity relative to the Earth is Vy_airplane.

Using the Pythagorean theorem, we can find the magnitude of the airplane's velocity relative to the Earth:

Magnitude of the airplane's velocity relative to Earth = sqrt((east-west component)^2 + (north-south component)^2)

= sqrt((580 km/h)^2 + (Vy_airplane)^2)

Now, we need to find Vy_airplane. We can do this by using trigonometry since the given information asks for the speed in terms of degrees south of east.

Since the airplane is moving eastward and there is no wind affecting the northward movement, the angle between the east direction and the airplane's velocity relative to Earth is 0 degrees.

Using trigonometry, we can say:

cos(0 degrees) = east-west component / magnitude of the airplane's velocity relative to Earth

cos(0 degrees) = 580 km/h / magnitude of the airplane's velocity relative to Earth

cos(0 degrees) = 1

Therefore, 1 = 580 km/h / magnitude of the airplane's velocity relative to Earth

Simplifying, we find:

Magnitude of the airplane's velocity relative to Earth = 580 km/h

So the speed of the airplane relative to Earth (magnitude of the velocity) is 580 km/h, and the direction is degrees south of east (since there is no northward component in the airplane's velocity relative to Earth).

To find the speed of the airplane relative to earth, we can use vector addition.

Let's represent the velocity of the airplane relative to the moving air as V_airplane, and the velocity of the wind as V_wind.

Given:
V_airplane = 580 km/h (easterly direction)
V_wind = 92 km/h (northward direction)

To find the velocity of the airplane relative to earth (V_airplane_earth), we need to add the vectors V_airplane and V_wind. Since they are at right angles to each other, we can use the Pythagorean theorem:

V_airplane_earth^2 = V_airplane^2 + V_wind^2

Substituting the given values:
V_airplane_earth^2 = (580 km/h)^2 + (92 km/h)^2

Calculating:
V_airplane_earth^2 = 336400 km^2/h^2 + 8464 km^2/h^2
V_airplane_earth^2 = 344864 km^2/h^2

Taking the square root of both sides:
V_airplane_earth ≈ 586.95 km/h

So, the speed of the airplane relative to earth is approximately 586.95 km/h.

To find the angle in degrees south of east, we can use trigonometry. Let's call the angle θ.

The component of V_wind that contributes to the southward motion is V_wind * sin(90 degrees). Since sin(90 degrees) equals 1, this component is simply V_wind itself.

θ = arctan(V_wind / V_airplane)

Substituting the given values:
θ = arctan(92 km/h / 580 km/h)
θ ≈ arctan(0.1586)

Calculating:
θ ≈ 9.12 degrees

Therefore, the speed of the airplane relative to earth is approximately 586.95 km/h, and the angle is approximately 9.12 degrees south of east.

X = 580 km/h = Velocity of airplane.

Y = 92 km/h = Velocity of wind.

Tan A = Y/X = 92/580 = 0.15862
A = 9.0o N. of E.

Speed = Sqrt(X^2+Y^2) = km/h.

Since the wind is northward, the
direction is N. of E. NOT S. of E.