An aeroplane flies 520 km north 45 degrees west then 387 km east . it is then noted that the plane is due north of its starting point. how far is the plane from its starting point?
How do I work it out?
a^2= 520^2-387^2
a^2= 270400-149769
a^2= 120631
a=347.32
Hope you made a sketch.
In mine I can use the cosine law
let the distance from the starting point be x
x^2 = 520^2 + 387^2 - 2(520)(387)cos45°
= 135572.6627
x = √....
= 368.2 km
To solve this problem, we can break it down into a step-by-step approach.
Step 1: Draw a diagram
Draw a diagram to represent the situation. Start with a reference point as the starting point and draw a line indicating the plane's initial direction of 520 km, 45 degrees west.
N
|
W--X--> E
|
S
Step 2: Find the vertical and horizontal components
Next, we need to find the vertical (north/south) and horizontal (east/west) components of the plane's movement.
The plane initially flies 520 km in the north 45 degrees west direction. We can use trigonometry to find the vertical and horizontal components of this movement.
Vertical Component = 520 km * sin(45°)
= 520 km * (√2 / 2)
= 520 km * 0.7071
≈ 368.30 km
Horizontal Component = 520 km * cos(45°)
= 520 km * (√2 / 2)
= 520 km * 0.7071
≈ 368.30 km
Step 3: Calculate the final position
Next, we will calculate the final position of the plane after flying 387 km east from the initial point.
Since the plane flies directly east, there is no change in its vertical component. The horizontal component increases by 387 km.
Final Vertical Position = 368.30 km
Final Horizontal Position = 368.30 km + 387 km
= 755.30 km
Step 4: Find the distance from the starting point
The distance from the starting point to the final position can be found using the Pythagorean theorem.
Distance = √((Final Vertical Position)^2 + (Final Horizontal Position)^2)
= √((368.30 km)^2 + (755.30 km)^2)
≈ 840.21 km
Therefore, the plane is approximately 840.21 km from its starting point.
To solve this problem, we can use vector addition and trigonometry. Let's break down the steps:
Step 1: Visualize the path
Draw a diagram to represent the path of the airplane. Start at the origin (representing the starting point) and move 520 km north 45 degrees west, then move 387 km east. The final position should be directly north of the starting point.
Step 2: Convert the vectors to Cartesian coordinates
Break down the movement of the airplane into two separate vectors: one for the north-west movement and one for the eastward movement.
For the north-west movement, we can use trigonometry to convert the polar coordinates (distance and angle) to Cartesian coordinates (x and y coordinates). Since the airplane moves 520 km north at an angle of 45 degrees west, we can calculate the x and y components as follows:
y = 520 km * sin(45 degrees)
x = 520 km * cos(45 degrees)
For the eastward movement, the y-component will be zero since it's moving strictly east.
Step 3: Add the vectors
Add the x and y components of the north-west movement vector to the corresponding components of the eastward movement vector.
Final x-coordinate = eastward movement's x-coordinate
Final y-coordinate = north-west movement's y-coordinate + eastward movement's y-coordinate
Step 4: Calculate the distance from the origin
Use the Pythagorean theorem to calculate the distance from the origin to the final position of the airplane:
Distance = sqrt((Final x-coordinate)^2 + (Final y-coordinate)^2)
By following these steps, you can work out the distance of the airplane from its starting point.