a commuter plane starts from an airport and takes the route whereby it starts to city,located 175km away in a directiin 30egrees north of east of east. Next, it flies for 150 km 20degrees west of north,to city b.finally, the plane flies 190kmdue west,to city c. Find the location of city c relative to the location of the starting point.
Add the three displacement vectors, to get the net displacement, D1 + D2 + D3.
D1 = 151.6 i + 87.5 j
D2 = -51.3 i + 141.0 j
D3 = -190 i
i is a unit vector East and j is a unit vector North.
To find the location of city C relative to the starting point, we need to break down the vectors and combine them together.
Let's start by drawing a diagram to visualize the information given.
- Start at the airport (point A)
- City A is located 175 km away in a direction 30 degrees north of east.
- Next, the plane flies for 150 km in a direction 20 degrees west of north to City B.
- Finally, the plane flies 190 km due west to City C.
To determine the overall displacement, we will break it down into x and y components:
1. Find the displacement from the airport (A) to City A:
- East component (x-direction): 175 km * cos(30°)
- North component (y-direction): 175 km * sin(30°)
2. Find the displacement from City A to City B:
- North component (y-direction): 150 km * cos(20°)
- West component (x-direction): 150 km * sin(20°)
3. Find the displacement from City B to City C:
- West component (x-direction): 190 km
4. Combine the x and y components of all three displacements:
- X-component: (175 km * cos(30°)) + (150 km * sin(20°)) + 190 km
- Y-component: (175 km * sin(30°)) + (150 km * cos(20°))
Finally, we can use these x and y components to find the magnitude and direction of the overall displacement from the starting point to City C using the Pythagorean theorem and trigonometry:
- Magnitude: sqrt((X-component)^2 + (Y-component)^2)
- Direction: tan^(-1)(Y-component / X-component)
So, by using the given distances and directions, we can calculate the relative location of City C from the starting point.