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?
d = 300km[135o]
V1 = 300km[135o]/0.75h = 400 km/h[135o]
without wind.
Vp + Vw = 400o[135]
Vp + 80[-90o] = 400[135]
Vp = 400[135] - 80[-90]
Vp=400*cos135+i400*sin135 - (80*Cos(-90)+i80*sin(-90))
Vp = -282.84+i282.84 - (0-i80)
Vp = -282.84 + i362.84 = 460km[128o]
= Velocity and heading of plane.
To determine the airspeed and heading of the plane, we need to consider the effect of the wind on its motion. Let's break down the problem and solve it step by step:
Step 1: Find the ground speed of the plane.
The ground speed is the speed at which the plane is moving relative to the ground. We can calculate it by adding the speed of the wind vector to the airspeed vector of the plane.
Given:
- Distance between City A and City B = 300 km
- Time to complete the trip = 0.75 hours
- Wind speed = 80 km/h (from the north)
Since the plane is flying from northwest (NW), we can think of its airspeed as having two components: one towards the north and one towards the west. Let's call the northward component "Vn" and the westward component "Vw."
Step 2: Solve for the ground speed components.
We can use the Pythagorean theorem to find the magnitude of the ground speed (Vg) of the plane:
Vg^2 = Vn^2 + Vw^2
Since the plane is flying at a constant speed, we can determine its airspeed components (Vn and Vw) by dividing the distance between City A and B (300 km) by the time to complete the trip (0.75 hours):
Vn = 300 km / 0.75 h = 400 km/h (northward component)
Vw = 0 km/h (no westward component since the plane is not moving in the east-west direction)
Step 3: Find the actual ground speed.
To find the actual ground speed, we need to add the speed of the wind vector to the airspeed vector of the plane:
Vg = Va + Vw
Given that the wind speed (Vw) is 80 km/h from the north, and the airspeed in the northward direction (Vn) is 400 km/h, the actual ground speed (Vg) can be calculated as follows:
Vg = √(Vn^2 + Vw^2) = √(400^2 + 80^2) = √(160000 + 6400) = √166400 ≈ 408.98 km/h
So, the ground speed of the plane is approximately 408.98 km/h.
Step 4: Find the airspeed of the plane.
To find the airspeed (Va), we need to subtract the wind speed (Vw) from the ground speed (Vg):
Va = Vg - Vw = 408.98 km/h - 80 km/h = 328.98 km/h
Therefore, the airspeed of the plane should be approximately 328.98 km/h.
Step 5: Find the heading of the plane.
The heading of the plane is the direction in which it should point or the angle at which it should fly. Since the wind is coming from the north (upwards), the plane needs to point slightly to the west to compensate for the wind.
We can use trigonometry to find the heading. The angle between the northward component (Vn) and the ground speed (Vg) will give us the heading angle (θ).
θ = arctan(Vw / Vn) = arctan(0 / 400) = arctan(0) = 0 degrees
Therefore, the heading of the plane should be 0 degrees (or due north).
In summary, to complete the trip from City A to City B in 0.75 hours with an 80 km/h northward wind, the plane should have an airspeed of approximately 328.98 km/h and a heading of 0 degrees (northward).