An airplane flies 200 km due west from city A to city B and then 245 km in the direction of 36.0° north of west from city B to city C.
(a) In straight-line distance, how far is city C from city A?
To find the straight-line distance between city A and city C, we can use the concept of vector addition.
First, let's break down the distance and direction of the airplane's flight from city B to city C. The airplane flies 245 km in the direction of 36.0° north of west.
We can represent this distance and direction as a vector. The magnitude of the vector is 245 km, and the angle north of west is 36.0°.
To calculate the horizontal and vertical components of this vector, we can use trigonometry.
The horizontal component (x-coordinate) can be found using the cosine function:
Horizontal component = 245 km * cos(36.0°)
The vertical component (y-coordinate) can be found using the sine function:
Vertical component = 245 km * sin(36.0°)
Now, let's calculate the horizontal and vertical components:
Horizontal component = 245 km * cos(36.0°) ≈ 197.43752 km
Vertical component = 245 km * sin(36.0°) ≈ 145.747724 km
Next, let's add the horizontal and vertical components to get the total displacement from city B to city C.
Total displacement = √((Horizontal component)^2 + (Vertical component)^2)
Total displacement = √((197.43752 km)^2 + (145.747724 km)^2)
Total displacement ≈ √(38980.5806 km^2 + 21238.7088 km^2)
Total displacement ≈ √(60219.2894 km^2)
Total displacement ≈ 245.39634 km
Therefore, the straight-line distance between city A and city C is approximately 245.4 km.
To find the straight-line distance from city A to city C, we can use the concept of vector addition.
First, we need to break down the given displacements into their respective components.
For the displacement from city A to city B, we are given that it is 200 km due west. Since this is a straight line, the horizontal component (x) is 200 km (since it is in the west direction) and the vertical component (y) is 0 km (since it is in the east-west plane). So, the displacement from city A to city B can be represented as (200 km, 0 km).
For the displacement from city B to city C, we are given that it is 245 km in the direction of 36.0° north of west. To find the x and y components of this displacement, we can use trigonometry. The horizontal component (x) is given by 245 km * cos(36.0°) and the vertical component (y) is given by 245 km * sin(36.0°). So, the displacement from city B to city C can be represented as (245 km * cos(36.0°), 245 km * sin(36.0°)).
To find the straight-line distance from city A to city C, we can add the two displacements together by adding their respective x and y components:
(x_AtoC, y_AtoC) = (x_AB + x_BC, y_AB + y_BC)
Substituting the given values, we get:
(x_AtoC, y_AtoC) = (200 km + 245 km * cos(36.0°), 0 km + 245 km * sin(36.0°))
Now, to find the straight-line distance, we can use the Pythagorean theorem:
distance = sqrt((x_AtoC)^2 + (y_AtoC)^2)
So, the straight-line distance from city A to city C is:
distance = sqrt((200 km + 245 km * cos(36.0°))^2 + (0 km + 245 km * sin(36.0°))^2)
Evaluating this expression will give us the final answer.