Answer:
(a) 609∠273°
(b) 5.31 km far away from the desired point to the east
5.31 far away from the desired point to the north
Explanation:
(a) In polar coordinates, the speed of the airplane is the next vector 640∠270° (640 km/h is the modulus and 'heading due south' is equivalent to an angle of 270° between the speed vector and the positive x-axis). On its part, the speed of the wind is the next vector 70∠45°.
The vector 640∠270° can be decomposed into Vx (x-coordinate component) and Vy (y-coordinate component) as follows:
Vx1 = 640cos(270°) = 0
Vy1 = 640sin(270°) = -640
For 70∠45°:
Vx2 = 70cos(45°) = 31.82
Vy2 =70sin(45°) = 31.82
The velocity of the airplane (V3) is calculated as follows:
Vx3 = Vx1 + Vx2 = 0 + 31.82 = 31.82
Vy3 = Vy1 + Vy2 = -640 + 31.82 = -608.18
The magnitude of the speed is:
V3 = √(Vx3² + Vy3²)
V3 = √(31.82² + (-608.18)²) = 609 km/h
α = tan (Vy3/Vx3)
α = tan (-608/31.82) = 273° = -87°
(b) Given that:
speed = displacement/time
Then:
displacement = speed time
At 640 km/h and after 10 min (= 1/6 h)
displacement = 640 1/6 = 106.67 km
This is the desired displacement, that is, in south direction.
At 609 km/h and after 10 min (= 1/6 h)
displacement = 609 1/6 = 101.5 km
Real displacement, in east of south direction (forming an angle of -87° with the positive x-coordinate) . This point is at the following distance form the desired point:
cos 87° 101.5 = 5.31 km far away from the desired point to the east
106.67 - sin 87° * 101.5 = 5.31 far away from the desired point to the north
