A plane is capable of flying 200 km/h. There is a wind of 90 km/h from the east. The pilot flies in a northwest direction relative to the ground for three hours from city A to city B. Find the speed of the plane relative to the ground, the heading of the plane, and the distance between city A and city B.
To find the speed of the plane relative to the ground, we need to use vector addition. The speed of the plane relative to the ground is equal to the vector sum of the plane's airspeed and the wind vector.
Step 1: Calculate the components of the wind vector.
The wind is blowing from the east at 90 km/h. Since the plane is flying in a northwest direction, we need to break down the wind vector into its east-west (x) and north-south (y) components. Let's call the east-west component of the wind vector Wx and the north-south component Wy.
Wx = 90 km/h * cos(45°) = 63.64 km/h
Wy = 90 km/h * sin(45°) = 63.64 km/h
Therefore, the wind vector can be expressed as (Wx, Wy) = (63.64 km/h, 63.64 km/h).
Step 2: Calculate the components of the plane's airspeed.
The plane's airspeed is given as 200 km/h. Since the pilot is flying in a northwest direction, we need to break down the airspeed vector into its east-west and north-south components. Let's call the east-west component of the airspeed vector Px and the north-south component Py.
Px = 200 km/h * cos(315°) = 141.42 km/h
Py = 200 km/h * sin(315°) = -141.42 km/h
Therefore, the plane's airspeed vector can be expressed as (Px, Py) = (141.42 km/h, -141.42 km/h).
Step 3: Add the airspeed vector and wind vector.
To find the resultant vector of the plane's airspeed and the wind, we simply add the corresponding components together.
Resultant east-west component R_x = Px + Wx = 141.42 km/h + 63.64 km/h = 205.06 km/h (rounded to two decimal places)
Resultant north-south component R_y = Py + Wy = -141.42 km/h + 63.64 km/h = -77.78 km/h (rounded to two decimal places)
The resultant vector can be expressed as (R_x, R_y) = (205.06 km/h, -77.78 km/h).
Step 4: Calculate the speed of the plane relative to the ground.
The speed of the plane relative to the ground is the magnitude of the resultant vector.
Speed of the plane relative to the ground = √(R_x^2 + R_y^2) = √[ (205.06 km/h)^2 + (-77.78 km/h)^2 ] = √(42023.80 km^2/h^2 + 6052.08 km^2/h^2) = √(48075.88 km^2/h^2) = 219.26 km/h (rounded to two decimal places)
Therefore, the speed of the plane relative to the ground is 219.26 km/h.
Step 5: Calculate the heading of the plane.
The heading of the plane is the angle between the plane's airspeed vector and the positive x-axis. To find this angle, we can use trigonometry.
Heading angle = arctan(R_y / R_x) = arctan(-77.78 km/h / 205.06 km/h) = arctan(-0.379) = -21.48° (rounded to two decimal places)
Note that the heading angle is negative because it is measured clockwise from the positive x-axis.
Therefore, the heading of the plane is approximately -21.48°.
Step 6: Calculate the distance between city A and city B.
The distance traveled by the plane can be calculated by multiplying the speed of the plane relative to the ground by the time spent flying.
Distance = Speed * Time = 219.26 km/h * 3 hours = 657.78 km (rounded to two decimal places)
Therefore, the distance between city A and city B is approximately 657.78 km.