Answer :
The addition of vectors using the graphical method consists of drawing the vectors on graphical coordinates and joining them by the process linking head to tail
The distance and direction of town B are as follows;
- The distance of town B from the airport is approximately 379.18 km
- The direction of town B from the airport is approximately 59.5° north of east
The given details of the motion and location of the plane are;
The distance of town A from the airport, d₁ = 335 km
The direction of town A from the airport = 20.0° north of east
The distance of town B from town A, d₂ = 245 km
The direction of town B from town A = 30.0° west of north
Required:
To use the graphical methods to determine the distance and direction from town B to the airport
Assumption (method):
Let the origin of the coordinate plane (0, 0) be the location of the airport
Solution:
The x and y-coordinates of the point A, is given as follows;
[tex]Coordinates \ of \ A = A(335 \times cos(20.0^{\circ}), 335 \times sin(20.0^{\circ}))[/tex]
∴ The coordinates of A = A(314.8, 114.6)
The x and y-coordinates of the point B is found as follows;
[tex]Coordinates \ of \ B = B(335 \times cos(20.0^{\circ}) - 245 \times sin(30.0^{\circ}), 335 \times sin(20.0^{\circ}) + 245 \times cos(30.0^{\circ}))[/tex]
∴ The coordinates of B = B(192.3, 326.8)
With above coordinates, the location of town A and town B can be graphed
From the obtained graph, the distance of town B from the airport, d₃, is given as follows;
[tex]d_3 = \sqrt{ 192.3^2+ 326.8^2 } \approx 379.18[/tex]
The distance of town B from the airport, d₃ is approximately 379.18 km
The direction of town B from the airport:
The direction north of west of town B from the airport, θ, is given as follows;
[tex]\theta \approx arctan\left(\dfrac{326.8}{192.3} \right) \approx 59.5 ^{\circ}[/tex]
The direction north of east of town B from the airport, θ ≈ 59.5° north of east
Learn more about the graphical method here:
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