an airplane flies 200 km due west from city A and then 350 km in the direction of 32.0 degrees north of west from city B to city C.
a) in a straight-line distance, how far is city c from city a?
b) relative to city a, what direction is city c?
a. 32 deg North of West = 148 deg.,CCW.
X=hor. = -200 + 350cos148 = -496.82km.
Y = ver. = 350sin148 = 185.47km.
d = sqrt(X^2+Y^2),
d = sqrt((-496.82)^2 + (185.47)^2 =
530.3km.
b. tanA = Y / X = 185.47 / -496.82 = -0.37332.
A = -20.47 deg,CW. = 180 - 20.47 = 159.53 deg.,CCW. = 20.47 deg, North of West..
To find the straight-line distance between City A and City C, we can use the Pythagorean theorem.
a) The airplane first flies 200 km due west from City A, and then it flies 350 km in the direction of 32.0 degrees north of west. We can break this velocity into a westward (horizontal) and northward (vertical) component.
Westward component: 350 km * cos(32.0 degrees) ≈ 295.46 km
Northward component: 350 km * sin(32.0 degrees) ≈ 186.61 km
Now, we can use these components to find the straight-line distance between City A and City C:
Distance = √((Westward component)^2 + (Northward component)^2)
Distance = √((295.46 km)^2 + (186.61 km)^2)
Distance ≈ 351.6 km (rounded to one decimal place)
So, the distance between City A and City C is approximately 351.6 km.
b) To determine the direction of City C relative to City A, we need to find the angle between the straight-line connecting them and the westward (horizontal) direction.
Angle = arctan(Northward component / Westward component)
Angle = arctan(186.61 km / 295.46 km)
Angle ≈ 32.7 degrees (rounded to one decimal place)
Therefore, the direction of City C relative to City A is approximately 32.7 degrees north of west.