Bob has a helicopter and from the launch pad he flies the following path. First he travels from the launch pad a distance of 33 kilometers at heading 10 degrees West of North. Then he flies 18 kilometers heading 45 degrees East of South. After this he flies 14 kilometers heading 50 degrees East of South. Now he is ready to return to the launch pad. What is the displacement vector that he needs to take to travel directly to the launch pad, from his present location?(for the heading give the number of degrees North of East-your answer may be greater than 90 degrees?)
33km @ 90+10 = 33km @ 100 deg.,CCW.
18km @ 270+45 = 18km @ 315 deg.,CCw.
14km @ 270+50 = 14km @ 320 deg.,CCW.
X=hor.=33cos100 + 18cos315 + 14cos320 = 5.73 + 12.73 + 10.72 = 17.7km.
Y=ver.=33sin100 + 18sin315 + 14sin320
= 2.95 + (-12.73) + 9.0 = -0.78km.
TanA = Y / X = -0.78 / 17.7 = -0.04407
A = -2.52 deg = 2.52 deg South of East.
D = -0.78 / sin(-2.52) = 17.67km @ 2.52
deg South of East.
The return path is 17.67km @ 2.52 deg
North of East.
To find the displacement vector that Bob needs to take to travel directly to the launch pad, we need to find the net displacement from his present location to the launch pad.
Let's break down Bob's path step by step:
1. First, Bob travels 33 kilometers at a heading of 10 degrees West of North. This means he is moving in a direction 10 degrees to the left of the North direction. We can represent this as a vector by splitting it into its North and East components. The North component can be calculated as 33 * sin(10°) and the East component can be calculated as 33 * cos(10°).
2. Next, Bob flies 18 kilometers at a heading of 45 degrees East of South. This means he is moving in a direction 45 degrees to the right of the South direction. We can represent this as a vector by splitting it into its South and East components. The South component can be calculated as 18 * sin(45°) and the East component can be calculated as 18 * cos(45°).
3. After that, Bob flies 14 kilometers at a heading of 50 degrees East of South. This means he is moving in a direction 50 degrees to the right of the South direction. Again, we can represent this as a vector by splitting it into its South and East components. The South component can be calculated as 14 * sin(50°) and the East component can be calculated as 14 * cos(50°).
Now, let's add up all the North, South, East, and West components to find the net displacement vector:
North Component: 33 * sin(10°)
South Component: -18 * sin(45°) - 14 * sin(50°)
East Component: 33 * cos(10°) + 18 * cos(45°) + 14 * cos(50°)
To get the heading of the net displacement vector, we can use the formula:
Heading = atan(East Component / North Component)
Finally, we can calculate the magnitude of the net displacement vector using the formula:
Magnitude = sqrt((East Component)^2 + (North Component)^2)
Let's calculate all the values:
North Component: 33 * sin(10°) ≈ 5.72 km
South Component: -18 * sin(45°) - 14 * sin(50°) ≈ -23.3 km
East Component: 33 * cos(10°) + 18 * cos(45°) + 14 * cos(50°) ≈ 4.34 km
Magnitude: sqrt((4.34 km)^2 + (5.72 km)^2) ≈ 7.11 km
Heading: atan(4.34 km / 5.72 km) ≈ 38.7°
Therefore, Bob needs to take a displacement vector of approximately 7.11 kilometers with a heading of 38.7 degrees North of East to travel directly to the launch pad from his present location.
To find the displacement vector that Bob needs to take to travel directly back to the launch pad, we need to calculate the net distance and net direction.
First, let's break down the distances and headings into their respective components:
1. The first leg has a distance of 33 kilometers and a heading 10 degrees West of North. To find the North and West components of this leg, we need to use trigonometry. Since the heading is West of North, we can subtract the angle from 90 degrees to get the angle from East, which is 80 degrees.
The North component can be found using the formula: North component = distance * sin(angle from East)
North component = 33 * sin(80) ≈ 32.82 kilometers
The West component can be found using the formula: West component = distance * cos(angle from East)
West component = 33 * cos(80) ≈ 6.76 kilometers
2. The second leg has a distance of 18 kilometers and a heading 45 degrees East of South. To find the South and East components of this leg, we need to again adjust the heading to be measured from East.
The South component can be found using the formula: South component = distance * sin(angle from East)
South component = 18 * sin(135) ≈ -12.73 kilometers (negative because it's South)
The East component can be found using the formula: East component = distance * cos(angle from East)
East component = 18 * cos(135) ≈ -12.73 kilometers (negative because it's East)
3. The third leg has a distance of 14 kilometers and a heading 50 degrees East of South. Again, we adjust the heading to be measured from East.
The South component can be found using the formula: South component = distance * sin(angle from East)
South component = 14 * sin(40) ≈ 8.98 kilometers
The East component can be found using the formula: East component = distance * cos(angle from East)
East component = 14 * cos(40) ≈ 10.68 kilometers
Now, let's calculate the net North and net East components by summing up the components from each leg:
Net North component = 32.82 kilometers
Net East component = 6.76 - 12.73 + 8.98 + 10.68 kilometers = 13.69 kilometers
Finally, we can find the displacement vector by combining the net North and net East components:
Displacement vector = √((Net North component)^2 + (Net East component)^2)
Displacement vector = √((32.82)^2 + (13.69)^2) ≈ 35.77 kilometers
Therefore, Bob needs to travel approximately 35.77 kilometers in a direction indicated by the angle whose tangent is (Net North component / Net East component) to return directly to the launch pad.