3. You travel from New Orleans, Louisiana, with the following vectors:
• New Orleans à Point A: q = 180°, Δd = 483 km
• Point A à Point B: q = 106°, Δd = 347 km
• Point B à Point C: q = 106°, Δd = 347 km
Remember: east = 0°, north = 90°, west = 180
Use the map to find your final destination. Where is the final destination, and what is its displacement from New Orleans? Show a data table containing each vector and its x and y components. Include the values of Δdnetx, Δdnety, Δdnet, tan qr, and qr.
To find the final destination and its displacement from New Orleans, you need to calculate the net displacement by summing up the individual displacements.
Let's start by calculating the x and y components of each vector using the given directions and magnitudes.
1. New Orleans to Point A:
• q = 180° (west)
• Δd = 483 km
The x-component of this vector is Δdx = -483 km (west) and the y-component is Δdy = 0 km (no movement in the north-south direction).
2. Point A to Point B:
• q = 106° (east of north)
• Δd = 347 km
To find the x and y components, we need to find the horizontal and vertical projections of this vector with respect to Point A. Using trigonometry:
Δdnetx = Δd * cos(q) = 347 km * cos(106°)
Δdnety = Δd * sin(q) = 347 km * sin(106°)
3. Point B to Point C:
• q = 106° (east of north)
• Δd = 347 km
The x and y components are the same as the previous vector since these vectors have the same direction and magnitude.
Now, let's calculate the x and y components for the second and third vectors, and also calculate the net displacement.
The x and y component table is as follows:
--------------------
| Vector | Δdx | Δdy |
--------------------
| 1 | -483 km| 0 km |
--------------------
| 2 | x₁ | y₁ |
--------------------
| 3 | x₂ | y₂ |
--------------------
To find x₁ and y₁, we can use the trigonometric functions:
x₁ = Δd * cos(q) = 347 km * cos(106°)
y₁ = Δd * sin(q) = 347 km * sin(106°)
Similarly, x₂ and y₂ are the same as x₁ and y₁.
Next, we can calculate the net displacement by adding up the x and y components:
Δdnetx = -483 km + x₁ + x₂
Δdnety = 0 km + y₁ + y₂
Finally, we can find the magnitude and direction of the net displacement using the formulas:
Δdnet = sqrt(Δdnetx^2 + Δdnety^2)
qr = atan(Δdnety / Δdnetx)
Now, you can substitute the values into the formulas above to calculate the net displacement and find its magnitude and direction.
To find the final destination and its displacement from New Orleans, we need to calculate the x and y components of each vector and then sum those components.
First, let's calculate the x and y components for each vector:
Vector from New Orleans to Point A:
q = 180°
Δd = 483 km
x component = Δd * cos(q) = 483 km * cos(180°) = -483 km
y component = Δd * sin(q) = 483 km * sin(180°) = 0 km
Vector from Point A to Point B:
q = 106°
Δd = 347 km
x component = Δd * cos(q) = 347 km * cos(106°) ≈ -107 km
y component = Δd * sin(q) = 347 km * sin(106°) ≈ 328 km
Vector from Point B to Point C:
q = 106°
Δd = 347 km
x component = Δd * cos(q) = 347 km * cos(106°) ≈ -107 km
y component = Δd * sin(q) = 347 km * sin(106°) ≈ 328 km
Now, let's calculate the net x and y components by summing the individual components:
Δdnetx = -483 km + (-107 km) + (-107 km) ≈ -697 km
Δdnety = 0 km + 328 km + 328 km ≈ 656 km
We can calculate the displacement magnitude Δdnet using Pythagorean theorem:
Δdnet = √(Δdnetx^2 + Δdnety^2) ≈ √((-697 km)^2 + (656 km)^2) ≈ √(484,809 km^2 + 430,336 km^2) ≈ √(915,145 km^2) ≈ 956.7 km
The displacement angle qr can be found using the arctan function:
tan qr = Δdnety / Δdnetx
qr = arctan(Δdnety / Δdnetx)
qr ≈ arctan(656 km / -697 km) ≈ -42.5°
Finally, we can locate the final destination by summing up the x and y components:
Final destination x coordinate = sum of x components = -483 km + (-107 km) + (-107 km) ≈ -697 km
Final destination y coordinate = sum of y components = 0 km + 328 km + 328 km ≈ 656 km
Therefore, the final destination is approximately 697 km west and 656 km north of New Orleans.