An air plane is flying due east relative to still air at a speed of 320 km/h. There is a 65 km/h wind blowing to ward the north, as measured by an observer on the ground. What is the velocity of the plane measured by the ground observer?
X = Vp = 320 km/h.
Y = Vw = 65km/h.
Vr = sqrt(X^2 + Y^2) = Resultant velocity.
vx=18.05 m/s and vy=88.88 m/s
To solve this problem, we need to find the resultant vector of the airplane's velocity and the velocity of the wind.
Step 1: Draw a diagram to represent the situation.
Draw a coordinate system with the x-axis representing east and the y-axis representing north.
Step 2: Identify the given values.
The speed of the airplane relative to still air is 320 km/h, and the speed of the wind blowing toward the north is 65 km/h.
Step 3: Break down the velocity vectors into their components.
The velocity of the airplane consists of a component in the x-direction (east) and a component in the y-direction (north). Let's call these components Vx and Vy, respectively. The velocity of the wind only has a component in the y-direction, which we'll call W.
Step 4: Determine the components of the vectors.
Since the airplane is flying due east, the x-component of its velocity is 320 km/h. The wind is blowing toward the north, so the y-component of its velocity is 65 km/h.
Vx = 320 km/h
Vy = 65 km/h
W = 65 km/h
Step 5: Find the resultant vector.
To find the resultant vector, we need to add the components of the airplane's velocity and the wind's velocity:
Rx = Vx
Ry = Vy + W
Rx = 320 km/h
Ry = 65 km/h + 65 km/h = 130 km/h
Step 6: Calculate the magnitude and direction of the resultant vector.
To find the magnitude of the resultant vector (R), we can use the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
R = sqrt((320 km/h)^2 + (130 km/h)^2)
R = sqrt(102400 km^2/h^2 + 16900 km^2/h^2)
R = sqrt(119300 km^2/h^2)
R ≈ 345.31 km/h
The direction of the resultant vector (θ) can be found using the inverse tangent function:
θ = arctan(Ry / Rx)
θ = arctan(130 km/h / 320 km/h)
θ ≈ 21.80°
Therefore, the velocity of the plane measured by the ground observer is approximately 345.31 km/h in the direction of 21.80° east of north.
To find the velocity of the plane measured by the ground observer, we can use vector addition.
First, let's represent the velocity of the plane relative to still air as a vector pointing east with a magnitude of 320 km/h.
Next, let's represent the velocity of the wind as a vector pointing north with a magnitude of 65 km/h.
To find the velocity of the plane measured by the ground observer, we need to add these two vectors. We can do this by considering the components of the vectors.
The velocity of the plane relative to still air can be represented as (320 km/h, 0 km/h) since it is only moving in the east direction.
The velocity of the wind can be represented as (0 km/h, 65 km/h) since it is only blowing in the north direction.
Now, we can add the components of these vectors to find the velocity of the plane measured by the ground observer.
Adding the east components: 320 km/h + 0 km/h = 320 km/h
Adding the north components: 0 km/h + 65 km/h = 65 km/h
Thus, the velocity of the plane measured by the ground observer is (320 km/h, 65 km/h).
Therefore, the plane is flying at an eastward speed of 320 km/h and a northward speed of 65 km/h, as measured by the ground observer.