A flight controller determines that an airplane is 20.0 south of him. Half an hour later, the same plane is 46.0 northwest of him.
If the plane is flying with constant velocity, what is the direction of its velocity during this time?
What is the magnitude of its velocity during this time?
If the controller is at (0,0) the plane moved from (0,-20) to (-32.53,32.53)
The distance is 61.79
Assuming the distances are in km, that would make the speed 123.58 km/hr
The direction is θ such that
tanθ = 52.53/-32.53
To determine the direction and magnitude of the plane's velocity, we can use the concept of average velocity. Average velocity is defined as the displacement divided by the time taken.
First, let's find the displacement of the airplane during this time. The displacement is the change in position from the starting point to the ending point.
Given that the airplane is initially 20.0 km south of the flight controller, and half an hour later, it is 46.0 km northwest of the flight controller.
To find the total displacement, we need to find the resultant of these two displacements. We can use vector addition for this purpose.
1. Draw a coordinate system on a piece of paper with north pointing up, south pointing down, east pointing towards the right, and west pointing towards the left.
2. Mark the initial position of the airplane 20.0 km south of the flight controller.
3. Draw an arrow pointing in the northwest direction with a magnitude of 46.0 km, starting from the initial position.
4. From the tail of the first arrow, draw an arrow pointing north with a magnitude of 20.0 km.
5. Connect the tail of the first arrow to the head of the second arrow with a straight line. This line represents the resultant displacement.
6. Measure the angle between the positive x-axis (east direction) and the line representing the resultant displacement. This angle represents the direction of the velocity.
7. Measure the length of the line representing the resultant displacement. This length represents the magnitude of the velocity.
Now, it's necessary to note that the northwest direction is the addition of the north and west components. So we need to find the north and west components of the 46.0 km vector.
Using trigonometry and the right triangle formed by the northwest vector with the north and west vectors, we can find the magnitudes of the north and west components.
Let's assume the north and west components are N and W, respectively.
Using the Pythagorean theorem, we have:
N^2 + W^2 = 46.0^2
Since the northwest component makes a 45-degree angle with the north and west components, we can use trigonometry to relate them:
tan(45°) = N/W => N = W
Substituting N = W into the first equation:
2N^2 = 46.0^2 => N^2 = 46.0^2 / 2 => N = W = 32.526 km
Now we have the north and west components, we can use these values to find the magnitude and direction of the displacement:
Magnitude of the displacement = √(N^2 + W^2) = √(32.526^2 + 32.526^2) = √(2 * 32.526^2) = √2 * 32.526 ≈ 45.98 km
To find the direction, we use the inverse tangent function:
Angle = tan^(-1) (N / W) = tan^(-1) (32.526 / 32.526) = 45°
So, we have found that the magnitude of the displacement of the airplane during this time is approximately 45.98 km and the direction is 45°.
Since the plane's velocity is constant, we can assume that the magnitude of the velocity is equal to the magnitude of the displacement, which is approximately 45.98 km.
Therefore, the direction of the plane's velocity during this time is 45°, and the magnitude of its velocity is approximately 45.98 km.