An airplane is flying due south with a velocity of 26.0 m/s.

a. If there is a tailwind of 27.0 m/s, what is the magnitude and direction of the actual velocity of the airplane? (a tailwind blows in the same direction as the motion of the airplane)
b. If there is a headwind of 27.0 m/s, what is the magnitude and direction of the actual velocity of the airplane? (a headwind blows across the direction and motion of the airplane)
c. If there is a crosswind of 27.0 m/s blowing due west, what is the magnitude and direction of the actual velocity of the airplane? (a crosswind blows across the direction and motion of the airplane)

26 m/s is about 10% from stalling for a Cessna 150.

With a tailwind, the airspeed does not change, but it will be added to the ground speed ("actual velocity").
With a headwind, the airspeed does not change either, but it will be subtracted from the ground speed.

Funny it may seem, with the headwind, the given aircraft is actually going backwards!

Wind is rarely in the same direction as the aircraft's velocity (as in (a) an (b)), so that in general, wind speed is added to the ground speed vectorially.
Airspeed: <0,-26>
wind speed: <-27,0>
ground speed: <-27,-26>
Magnitude: √(27²+26²)= 37.48 m/s
Direction (CCW from east):atan(-26/-27)=223.9°

To find the magnitude and direction of the actual velocity of the airplane in each scenario, we can use vector addition:

a) With a tailwind, the actual velocity can be found by adding the velocity of the airplane to the velocity of the tailwind.

The velocity of the airplane is given as 26.0 m/s due south.
The velocity of the tailwind is given as 27.0 m/s.

To represent these velocities as vectors, we can use the positive direction of the x-axis as east and the positive direction of the y-axis as north. Therefore, the velocity of the airplane is (0, -26.0) m/s (a negative value indicating south), and the velocity of the tailwind is (27.0, 0) m/s (only in the east direction).

Adding these vectors together, we get:
(0, -26.0) + (27.0, 0) = (27.0, -26.0) m/s.

The magnitude of the actual velocity can be found using the Pythagorean theorem:
Magnitude of the actual velocity = sqrt((27.0)^2 + (-26.0)^2) ≈ 36.87 m/s.

The direction of the actual velocity can be found using the inverse tangent function:
Direction of the actual velocity = arctan(-26.0/27.0) ≈ -46.79 degrees.

Therefore, in this scenario, the magnitude of the actual velocity of the airplane is approximately 36.87 m/s, and its direction is approximately 46.79 degrees south of east.

b) With a headwind, we subtract the velocity of the headwind from the velocity of the airplane.

The velocity of the airplane remains the same as before: (0, -26.0) m/s.

Now, the velocity of the headwind is given as 27.0 m/s in the west direction. Represented as a vector, it is (-27.0, 0) m/s.

Subtracting the vectors, we get:
(0, -26.0) - (27.0, 0) = (-27.0, -26.0) m/s.

The magnitude of the actual velocity can be found as before:
Magnitude of the actual velocity = sqrt((-27.0)^2 + (-26.0)^2) ≈ 36.87 m/s.

The direction of the actual velocity can be found using the inverse tangent function:
Direction of the actual velocity = arctan((-26.0)/(-27.0)) ≈ 44.42 degrees.

Therefore, in this scenario, the magnitude of the actual velocity of the airplane is approximately 36.87 m/s, and its direction is approximately 44.42 degrees south of west.

c) With a crosswind, we use vector addition again.

The velocity of the airplane remains the same as before: (0, -26.0) m/s.

Now, the velocity of the crosswind is given as 27.0 m/s in the west direction. Represented as a vector, it is (-27.0, 0) m/s.

Adding the vectors together, we get:
(0, -26.0) + (-27.0, 0) = (-27.0, -26.0) m/s.

The magnitude of the actual velocity can be found as before:
Magnitude of the actual velocity = sqrt((-27.0)^2 + (-26.0)^2) ≈ 36.87 m/s.

The direction of the actual velocity can be found using the inverse tangent function:
Direction of the actual velocity = arctan((-26.0)/(-27.0)) ≈ 44.42 degrees.

Therefore, in this scenario, the magnitude of the actual velocity of the airplane is approximately 36.87 m/s, and its direction is approximately 44.42 degrees south of west.