An airplane is heading due south at a speed of 640 km/h. if a wind begins blowing from the southwest at a speed of 70 km/h (average), calculate the following.
(a) the velocity (magnitude and direction) of the plane relative to the ground
643.81
km/h
° (east of south)
(b) how far from its intended position will it be after 10 min if the pilot takes no corrective action. [hint: first draw a diagram.]

(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 = 640*cos(270°) = 0

Vy1 = 640*sin(270°) = -640

For 70∠45°:

Vx2 = 70*cos(45°) = 31.82

Vy2 =70*sin(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

the answer is a. at the equator