An paris aeroplane left Benin airport at 1:00pm to fly to Yola airport 600km away on a bearing of 110• at an airspeed of 220km/hr. On that particular day, the meteorologists reported that there existed a steady wind of 35km/hr from a bearing of 40•. What is the course of the aeroplane and it's time of arrival.
wind components of speed:
north -35 cos 40 = -26.8
east -35 sin 40 = -22.5
heading angle clockwise from N = h
north air speed = 220 cos h
east air speed = 220 sin h
speed made good = v
direction = 110
cos 110 = -.342
sin 110 = .940
north = v cos 110 = -.342 v
east = v sin 110 = .940 v
so
-26.8 + 220 cos h = -.342 v
-22.5 + 220 sin h = +.940 v
eliminate v to get heading h
multiply first by 2.75
-73.7 + 605 cos h = -.940 v
-22.5 + 220 sin h = +.940 v add
--------------------------
- 96.2 + 605 cos h + 220 sin h = 0
The easy way is try angles until you converge but
605 cos h + 220 sin h = 96.2
let y = sin h
then cos h = sqrt(1-y^2)
605(1-y^2)^.5 + 220 y = 96.2
(1-y^2)^.5 + .364 y = .159
(1-y^2)^.5 = .159-.364 y
1 - y^2 = .0253 - .116 y + .132 y^2
1.132 y^2 - .116 y -.975 = 0
y = .983 or -.878
y = sin h so h = 79.4 or -61.4
since we want to end up going 20 deg south of east and we are being pushed south by wind we better pick that heading of 79.4 deg east of north
let's call it 80 degrees
then remember
-22.5 + 220 sin h = +.940 v
-22.5 + 217 = .940 v
v = 207 speed made good
600 km/207 km/h = 2.9 hours
1 pm + 2.9 hours ---> 4 pm about
To determine the course of the airplane and its time of arrival, we need to take into account the effect of wind on its trajectory. We can break this down into two components:
1. Ground speed (speed relative to the ground):
Ground speed is the speed at which the airplane is actually covering the distance along its intended course, taking into account the effect of wind. It can be calculated using the vector addition of the airplane's airspeed and the wind speed.
2. Heading (direction or course of the airplane):
Heading is the angle between the direction the airplane is pointing (its course) and true north. It can be calculated using the vector addition of the airplane's course and the wind direction.
Now let's calculate step-by-step:
Step 1: Calculate the ground speed:
Ground speed = airspeed - wind speed
Ground speed = 220 km/hr - 35 km/hr = 185 km/hr
Step 2: Calculate the heading:
Heading = course + wind direction
Heading = 110° + 40° = 150°
So, the airplane's ground speed is 185 km/hr, and its heading is 150°.
Step 3: Calculate the time of arrival:
Time = Distance / Speed
Time = 600 km / 185 km/hr ≈ 3.24 hours
Since the time of departure was 1:00 PM, we can calculate the time of arrival by adding 3.24 hours to it:
Step 4: Calculate the time of arrival:
Time of arrival = 1:00 PM + 3.24 hours
Time of arrival ≈ 4:14 PM
Therefore, the airplane's course is 150°, and it will arrive at approximately 4:14 PM.
To find the course of the airplane and its time of arrival, we can use vector addition. Let's break down the problem into two components: the airplane's velocity relative to the ground and the groundspeed and heading.
1. Airplane's velocity relative to the ground:
The airspeed of the airplane is given as 220 km/hr. However, there is also a wind of 35 km/hr that affects the airplane's velocity relative to the ground. We can consider the wind as a vector acting in the opposite direction of its bearing.
To calculate the horizontal (x-component) and vertical (y-component) of the airplane's velocity relative to the ground, we need to use trigonometry. Let's consider the horizontal component first:
Horizontal Component:
The horizontal component of the airspeed (220 km/hr) is given by:
Horizontal Airspeed = Airspeed * cos(Angle of Bearing)
Horizontal Airspeed = 220 km/hr * cos(110°)
Vertical Component:
The vertical component of the airspeed (220 km/hr) is given by:
Vertical Airspeed = Airspeed * sin(Angle of Bearing)
Vertical Airspeed = 220 km/hr * sin(110°)
2. Groundspeed and Heading:
The groundspeed is the magnitude of the airplane's velocity relative to the ground calculated in the previous step. The heading is the direction in which the airplane is moving relative to the ground.
To find the groundspeed, we can use the Pythagorean theorem:
Groundspeed = √(Horizontal Airspeed^2 + Vertical Airspeed^2)
To find the heading, we use the inverse tangent (arctan) function:
Heading = 180° + arctan(Vertical Airspeed / Horizontal Airspeed)
Now let's calculate the components:
Horizontal Airspeed = 220 km/hr * cos(110°)
Horizontal Airspeed ≈ -58.16 km/hr (Negative because it acts in the opposite direction of the bearing)
Vertical Airspeed = 220 km/hr * sin(110°)
Vertical Airspeed ≈ 208.91 km/hr
Groundspeed = √((-58.16 km/hr)^2 + (208.91 km/hr)^2)
Groundspeed ≈ 218.1 km/hr
Heading = 180° + arctan(208.91 km/hr / -58.16 km/hr)
Heading ≈ 178.1°
So, the course of the airplane is approximately 178.1°, and the groundspeed is approximately 218.1 km/hr.
To find the time of arrival, we can divide the distance between Benin and Yola airports (600 km) by the groundspeed:
Time of Arrival = Distance / Groundspeed
Time of Arrival = 600 km / 218.1 km/hr
Time of Arrival ≈ 2.75 hours (or 2 hours and 45 minutes)
Therefore, the airplane's course is approximately 178.1°, and it will arrive at Yola airport in approximately 2 hours and 45 minutes.