An airplane has a velocity of 400 km/h relative to the wind. At the same time, a wind blows northward with a speed of 75 km/h relative to earth. Find the direction that the airplane must head in order to move in an easterly direction relative to earth.
Wouldn't it need to fly in a direction that would cancel the North drift?
400sinTheta=75 where theta is the angle S of E.
is there any answer
To find the direction that the airplane must head in order to move in an easterly direction relative to earth, we need to analyze the vectors involved.
Let's assume that the direction of the wind blowing northward is positive, and the direction of the airplane's velocity relative to the wind is also positive.
Given:
Airplane velocity relative to wind = 400 km/h
Wind speed = 75 km/h
We can treat the airplane's velocity relative to the earth as the vector sum of its velocity relative to the wind and the wind velocity.
Mathematically, we can write this as:
(Airplane velocity relative to earth)^2 = (Airplane velocity relative to wind)^2 + (Wind velocity)^2 + 2 * (Airplane velocity relative to wind) * (Wind velocity) * cosθ
where θ is the angle between the airplane's velocity relative to the wind and the wind velocity.
Since we want the airplane to move in an easterly direction relative to earth, the net velocity (Airplane velocity relative to earth) should have an easterly component.
This means that the angle θ between the airplane's velocity relative to the wind and the wind velocity should be between 0 degrees and 90 degrees, inclusive.
Simplifying the equation further:
(Airplane velocity relative to earth)^2 = (400 km/h)^2 + (75 km/h)^2 + 2 * (400 km/h) * (75 km/h) * cosθ
Now we can solve for the angle θ.
Let's substitute the given values into the equation:
(Airplane velocity relative to earth)^2 = (400 km/h)^2 + (75 km/h)^2 + 2 * (400 km/h) * (75 km/h) * cosθ
(Airplane velocity relative to earth)^2 = 160000 km^2/h^2 + 5625 km^2/h^2 + 60000 km^2/h^2 * cosθ
(Airplane velocity relative to earth)^2 = 221625 km^2/h^2 + 60000 km^2/h^2 * cosθ
Knowing that (Airplane velocity relative to earth) > 0, we can solve for cosθ:
cosθ = [(Airplane velocity relative to earth)^2 - 221625 km^2/h^2] / (60000 km^2/h^2)
cosθ = [(Airplane velocity relative to earth)^2 - 221625 km^2/h^2] / 60000 km^2/h^2
Using a calculator, we can find the cosine inverse of the above value to determine the angle θ between the airplane's velocity relative to the wind and the wind velocity.
Once we have obtained the angle θ, we can determine the direction that the airplane must head by adding or subtracting this angle from the direction of the wind.
Since the wind is blowing northward, and we want the airplane to move in an easterly direction, the direction the airplane should head is east of north.
Note: The exact calculations for θ and the resulting direction will depend on the specific values plugged into the equation, which can be found using a calculator.
To find the direction that the airplane must head in order to move in an easterly direction relative to the Earth, we need to consider the velocities of the airplane and the wind.
Let's break down the problem:
1. The velocity of the airplane relative to the wind is given as 400 km/h.
2. The velocity of the wind relative to the Earth is given as 75 km/h northward.
Now, let's analyze the situation. Since we want the airplane to move in an easterly direction relative to the Earth, we need to take into account the vectors involved.
The vector sum of the airplane's velocity and the wind's velocity will give us the resultant velocity, which is the velocity of the airplane relative to the Earth.
Therefore, we can calculate the resultant velocity using vector addition.
To do that, we need to take into account both the magnitude (speed) and direction of the velocities.
1. Start by drawing a vector representing the velocity of the airplane 400 km/h in some direction. Let's assume it's to the north-east.
2. Then, draw another vector representing the velocity of the wind 75 km/h northward.
3. Since we want the airplane to move in an easterly direction relative to the Earth, we need to find a resultant vector that has a more easterly component.
To find this resultant vector, we need to add the airplane's velocity vector and the wind's velocity vector together using vector addition.
Now, let's calculate the magnitude and direction of the vector:
Using vector addition, the magnitude of the resultant vector can be found using the Pythagorean theorem:
Resultant magnitude = √[(airplane velocity magnitude)² + (wind velocity magnitude)²]
Resultant magnitude = √[(400 km/h)² + (75 km/h)²]
Resultant magnitude ≈ 408.28 km/h
The direction of the resultant vector can be found using the inverse tangent function:
Resultant direction = arctan(wind velocity magnitude/airplane velocity magnitude)
Resultant direction ≈ arctan(75 km/h / 400 km/h)
Resultant direction ≈ arctan(0.1875)
Resultant direction ≈ 10.67 degrees north of east
Therefore, to move in an easterly direction relative to the Earth, the airplane must head approximately 10.67 degrees north of east.