A pilot is flying from City A to City B which is 300 km [NW]. If the plane will encounter a constant wind of 80 km/h from the north and the schedule insists that he complete his trip in 0.75 h, what air speed and heading should the plane have?

Vp = 300km[135o]/0.75h = 400 km/h[135o]

Without the wind.

Vp + Vw = 400km/h[135] = -282.84+282.84i
Vp - 80i = -282.84 + 282.84i
Vp = -282.84 + 282.84i + 80i

Vp = -282.84 + 362.8i=460.4km/h[128o] = 460.4km/h[38o] W. of N. With the wind
blowing.

To determine the air speed and heading of the plane, we need to consider the effect of the wind on the plane's motion.

Given:
- Distance between City A and City B (d) = 300 km
- Wind speed (v_wind) = 80 km/h (from north)
- Time to complete the trip (t) = 0.75 hours

Let's first understand the effect of the wind on the plane's motion. The wind will create a force called the "wind vector" acting against the motion of the plane. Since the wind is coming from the north (opposite to the direction of City B), we can represent this force as a vector pointing south with a magnitude of 80 km/h.

Now, let's assume that the plane is traveling at a certain airspeed (v_air) in a direction called the "heading." We want to determine the airspeed and heading that will allow the plane to overcome the wind and reach City B in the given time.

To solve this problem, we need to split the plane's motion into two components: the horizontal component (parallel to the ground) and the vertical component (perpendicular to the ground). The vertical component is affected by the wind, while the horizontal component represents the plane's forward motion.

Let's assume the plane's airspeed is represented by v_x and v_y for the horizontal and vertical components, respectively. Since the plane's motion is in the northwest direction, we can express the horizontal and vertical components as follows:

v_x = v_air * cos(θ)
v_y = v_air * sin(θ)

where θ is the angle between the plane's heading and the positive x-axis (east).

Now, let's consider the horizontal (east-west) motion of the plane. The time taken to cover the horizontal distance (d_x) between City A and City B can be calculated as:

t = d_x / v_x

Since the time is given as 0.75 hours and the distance (d_x) is 300 km, we have:

0.75 = 300 / v_x

Solving for v_x, we find:

v_x = 400 km/h

Now, let's consider the vertical (north-south) motion of the plane. The vertical component of the wind vector (v_wind_y) opposes the plane's vertical motion. Therefore, the net vertical velocity (v_net_y) will be:

v_net_y = v_y - v_wind_y

The vertical distance covered by the plane (d_y) can be calculated as:

d_y = v_net_y * t

We know that the distance covered vertically (d_y) is zero since the plane is flying from City A to City B. Therefore, we have:

0 = v_net_y * t

Substituting the values, we find:

0 = v_air * sin(θ) - v_wind * sin(180°)

Since sin(180°) = 0, we can solve for θ:

v_air * sin(θ) = v_wind * sin(180°)
v_air * sin(θ) = v_wind * 0
v_air * sin(θ) = 0

To satisfy this equation, sin(θ) must be zero. Therefore, θ must be either 0° or 180°. However, since the wind is coming from the north, it opposes the plane's motion. Thus, θ should be:

θ = 180°

Now that we know the angle, we can calculate the airspeed using the horizontal component equation:

v_air = v_x / cos(θ)
v_air = 400 km/h / cos(180°)
v_air = 400 km/h / (-1)
v_air = -400 km/h

The airspeed of the plane should be 400 km/h, heading south (180°) to counteract the north wind and maintain a course towards City B.