Bob has a helicopter and from the launch pad he flies the following path: First he travels from the launch pad a distance of 24 kilometers at heading of 60 degrees East of South. Then he flies 48 kilometers heading 75 degrees West of North. After this he flies 35 kilometers heading 77 West of North. Now he is ready to return directly to the launch pad. What is the displacement vector that he needs to take to return directly to the launch pad from his present location (for the heading give the number degrees north of east - your answer may be greater than 90 degrees)?
24@S60°E = (24sin60°,-24cos60°)
48@n75°W = (-48sin75°,48cos75°)
Now just add them up, and convert back to compass heading.
what about 35 km @ 77 degrees ? do i just do 35@n77°W = ( -35sin77°, 35cos77° )? or did you already add it into the other north of west?
and also, which ones should i be adding together? the x's together and y's together?
See previous post: Sun, 10-8-17, 10:47PM.
To find the displacement vector that Bob needs to take to return directly to the launch pad, we can break down his journey into horizontal and vertical components.
1. Start by calculating the horizontal and vertical components of each leg of Bob's journey:
First leg:
- Distance: 24 km
- Heading: 60 degrees East of South
- Vertical Component: 24 km * sin(60) = 24 * (√3/2) = 12√3 km South
- Horizontal Component: 24 km * cos(60) = 24 * (1/2) = 12 km East
Second leg:
- Distance: 48 km
- Heading: 75 degrees West of North
- Vertical Component: 48 km * sin(75) = 48 * (√3 - 1)/2 ≈ 20.53 km North
- Horizontal Component: 48 km * cos(75) = 48 * (√3 + 1)/2 ≈ 37.70 km West
Third leg:
- Distance: 35 km
- Heading: 77 degrees West of North
- Vertical Component: 35 km * sin(77) = 35 * (√3 + 1)/2 ≈ 27.76 km North
- Horizontal Component: 35 km * cos(77) = 35 * (√3 - 1)/2 ≈ 10.67 km West
2. Calculate the total horizontal and vertical components by adding up the individual components:
Total Horizontal Component: 12 km (East) - 37.70 km (West) - 10.67 km (West) = -36.37 km
Total Vertical Component: 12√3 km (South) + 20.53 km (North) + 27.76 km (North) ≈ 20.53√3 km
3. Convert the total horizontal and vertical components to a single displacement vector:
Use the Pythagorean theorem to find the magnitude of the displacement vector:
Magnitude: √((-36.37 km)^2 + (20.53√3 km)^2) ≈ 41.82 km
Use trigonometry to find the direction (heading) of the displacement vector:
Direction: arctan((20.53√3 km) / (-36.37 km)) ≈ -62.05 degrees (measured counterclockwise from the positive x-axis)
Therefore, Bob needs to take a displacement vector of approximately 41.82 km magnitude, oriented at a heading of -62.05 degrees (measured counterclockwise from the positive x-axis) to return directly to the launch pad.