A plane flies due north (90° from east) with a velocity of 100 km/h for 3 hours. During this time, a steady wind blows southeast at 30 km/h at an angle of 315° from due east. After 3 hours, where will the plane’s position be relative to its starting point?
PLEASE explain what this 315° angle means? This is a problem related to triangles and a triangle can only have 180° for total angles. I assumed this meant that I would do 315-270 to get the difference between the axis and the angle, which results in 45°. However, everything online says otherwise but they gloss over the fact that a triangle can't be more than 180°.
they use the unit circle ... 360º ... E (0º), N (90º), W (180º), S (270º)
so the wind is pushing the plane south and east
after 3 hr
... north ... 3[100 - (30 / √2)]
... east ... 3 (30 / √2)
I believe the question is asking for the displacement distance in kilometers, and the angle of change from the starting point?
use the Pythagorean theorem, and note that
tan(90-θ) = y/x
The angle of 315° is actually measured counterclockwise from the positive x-axis (east) to the direction of the wind. In this case, 315° is equivalent to 45° clockwise from the positive x-axis (east).
You are correct that a triangle can't have more than 180° in its angles. However, in this problem, the 315° angle is not related to the triangle formed by the plane's path and the wind direction. Instead, it is used to determine the wind's direction relative to the east axis.
To find the position of the plane after 3 hours, we need to break down the velocities into their components. First, let's analyze the plane's velocity.
The plane is flying due north with a velocity of 100 km/h. Since it is traveling directly north, it has no velocity component in the east direction. Therefore, the plane's velocity components are:
Vx (eastward component) = 0 km/h
Vy (northward component) = 100 km/h
Next, let's analyze the wind's velocity.
The wind is blowing southeast at an angle of 315° from due east and with a magnitude of 30 km/h. To find the wind's velocity components, we can use trigonometry.
The wind's eastward component (Wx) can be found using the cosine function:
Wx = magnitude * cos(angle)
Wx = 30 km/h * cos(315°)
= 30 km/h * cos(45°) (since cos(315°) = cos(45°) due to symmetry)
= 30 km/h * (√2/2)
= 15√2 km/h
The wind's northward component (Wy) can be found using the sine function:
Wy = magnitude * sin(angle)
Wy = 30 km/h * sin(315°)
= 30 km/h * sin(45°) (since sin(315°) = sin(45°) due to symmetry)
= 30 km/h * (√2/2)
= 15√2 km/h
Therefore, the wind's velocity components are:
Wx (eastward component) = 15√2 km/h
Wy (northward component) = 15√2 km/h
Now, let's find the resultant velocity of the plane plus wind:
Rx = Vx + Wx = 0 km/h + 15√2 km/h = 15√2 km/h
Ry = Vy + Wy = 100 km/h + 15√2 km/h = 100 km/h + 15√2 km/h
Since the plane flies for 3 hours, we can multiply the resultant velocity components by 3:
Rx(final) = Rx * 3 = 15√2 km/h * 3 = 45√2 km
Ry(final) = Ry * 3 = (100 km/h + 15√2 km/h) * 3 = 300 km/h + 45√2 km/h
Therefore, after 3 hours, the plane's position relative to its starting point will be 45√2 km east and 300 km north.