Answer:
a) v = (-338.1î + 725ĵ) km/h
b) Velocity of the wind = (-838.1î + 725ĵ) km/h
Magnitude of the wind = 1108.17 km/h
Direction = -40.9°; that is, 40.9° in the clockwise direction from the positive x-axis.
Step-by-step explanation:
a) Airplane is flying 800 km/h in a direction 25° west of North.
Velocity of the plane = (vₓ, vᵧ)
vₓ = v cos θ
vᵧ = v sin θ
v = magnitude of the velocity = 800 km/h
θ = angle the velocity makes with the positive x-axis = 90° + 25° = 115°
vₓ = v cos θ
vₓ = 800 cos 115° = - 338.1 km/h
vᵧ = v sin θ
vᵧ = 800 sin 115° = 725 km/h
v = (-338.1î + 725ĵ) km/h
b) What speed and direction should the wind be in order for the airplane to now fly 500 km/h due east
Relative velocity of airplane with respect to the wind
= (velocity of the airplane) - (velocity of the wind)
Note that the velocities on the right hand side are with respect to earth's frame of reference.
Relative velocity of airplane with respect to the wind = 500 km/h east = (500î) km/h
velocity of the airplane = (-338.1î + 725ĵ) km/h
velocity of the wind = ?
500î = (-338.1î + 725ĵ) - (velocity of the wind)
(velocity of the wind) = (-338.1î + 725ĵ) - 500î
= (-838.1î + 725ĵ) km/h
Velocity of the wind = (-838.1î + 725ĵ) km/h
Magnitude = √[(-838.1)² + (725²)] = 1108.17 km/h
Direction = tan⁻¹ (725 ÷ -838.1) = -40.9°
Hope this Helps!!!