An aircraft travels at 250km/h in still air. If a wind of 50km/h blows steadily from the southwest and the pilot wishes to fly due north, find the course he must set and the ground speed of the aircraft. On the return journey the wind speed and direction are unaltered. Find the new course and ground speed.
As with all geometry and related problems, the first step is to draw a triangle of velocities.
Wind is from south-west at 50 km/h.
Draw a line from A to B with an arrow pointing north-east. From B, draw a line (approx. 5 times as long) towards north, ending at C which is exactly north of A.
So we have an oblique triangle ABC, with angle A=45 degrees, AB=50, BC=250.
Solve by sine-rule,
sin(A)/BC=sin(C)/AB, or
sin(C)=sin(A)*AB/BC
=sqrt(2)/2*(50/250)
=.14142
and =>
A=8.1301°
and
B=180-(A+C)=126.8699°
Hence apply the sine rule again to get
AC=sin(B)*BC/sin(A)
=sin(126.8699)*250/sin(45)
=282.843 km
You can proceed similarly for the return trip, and I'll leave that to you. If you encounter difficulties, feel free to post what you have done and we'll take it from there.
Well, well, well, looks like our pilot is embarking on a windy adventure! Alright, let's tackle this question with a dash of clown humor.
To find the course he must set, the pilot needs to take into account the wind's speed and direction. Since the wind is blowing from the southwest, it is hitting the aircraft at an angle of 45 degrees to the north. So, to counteract the wind and fly directly north, our pilot would need to set a course of 45 degrees to the east of north, essentially pointing his nose slightly to the right. Be careful not to tickle the clouds, pilot!
Now, onto the ground speed. The ground speed is the speed at which the aircraft moves relative to the ground. To calculate this, we need to take into account the effect of the wind.
Since the wind is blowing from the southwest at 50 km/h, it is opposing the aircraft's northward motion. The component of the wind that opposes the aircraft's direction is given by 50 * sin(45 degrees), which equals 35.4 km/h.
Therefore, the ground speed of the aircraft would be the difference between its still air speed (250 km/h) and the opposing component of the wind (35.4 km/h), which equals 214.6 km/h. Our pilot is making good progress despite the winds!
Now, on the return journey, the wind speed and direction remain the same. Lovely consistency, don't you think? So, to find the new course, the pilot simply needs to adjust by 45 degrees to the west of north, pointing his nose slightly to the left this time. No worries, pilot, clown bot has got your compass covered!
As for the new ground speed, we follow the same process as before. The opposing component of the wind (50 * sin(45 degrees)) remains 35.4 km/h, and subtracting this from the still air speed (250 km/h) gives us a new ground speed of 214.6 km/h. It seems like the wind didn't blow the clown out of our pilot's speedometer!
So there you have it, dear friend! Our pilot sets a course of 45 degrees east of north, with a ground speed of 214.6 km/h. And on the return journey, the course is adjusted to 45 degrees west of north, with the same ground speed. Safe travels, happy flying, and may the winds always be in your favor! 🌬️✈️
To find the course the pilot must set and the ground speed of the aircraft, we need to apply vector addition.
1. Flight to the north:
The aircraft wants to fly due north, but there is a wind blowing from the southwest. We can break down the aircraft's velocity and the wind's velocity into their components.
Aircraft's velocity (in still air):
- Speed = 250 km/h
- Direction = Due north (0°)
Wind velocity:
- Speed = 50 km/h
- Direction = Southwest (225°)
To find the course the pilot must set, we need to find the resultant vector by adding the aircraft's velocity vector and the wind's velocity vector.
X-component of aircraft's velocity = Speed * cos(Direction) = 250 km/h * cos(0°) = 250 km/h
Y-component of wind velocity = Speed * sin(Direction) = 50 km/h * sin(225°) = -35.355 km/h
X-component of resultant vector = X-component of aircraft's velocity + X-component of wind velocity = 250 km/h + 0 km/h = 250 km/h
Y-component of resultant vector = Y-component of aircraft's velocity + Y-component of wind velocity = 0 km/h + (-35.355 km/h) = -35.355 km/h
The resultant vector has a negative Y-component, indicating that the pilot needs to set the course slightly east of north to counteract the southwest wind.
Course the pilot must set = arctan(Y-component / X-component) = arctan((-35.355 km/h) / 250 km/h) ≈ -8.152° (east of north)
Ground speed of the aircraft is the magnitude of the resultant vector:
Magnitude of resultant vector = sqrt((X-component)^2 + (Y-component)^2) = sqrt((250 km/h)^2 + (-35.355 km/h)^2) ≈ 252.28 km/h
Therefore, the pilot must set a course approximately 8.152° east of north, and the ground speed of the aircraft will be approximately 252.28 km/h.
2. Return journey:
The wind speed and direction remain unaltered, but the pilot is flying back from the north to the south.
To find the new course the pilot must set and the ground speed, we apply the same process as above.
X-component of aircraft's velocity = Speed * cos(Direction) = 250 km/h * cos(180°) = -250 km/h
Y-component of wind velocity = Speed * sin(Direction) = 50 km/h * sin(225°) = -35.355 km/h
X-component of resultant vector = X-component of aircraft's velocity + X-component of wind velocity = -250 km/h + 0 km/h = -250 km/h
Y-component of resultant vector = Y-component of aircraft's velocity + Y-component of wind velocity = 0 km/h + (-35.355 km/h) = -35.355 km/h
The resultant vector has a negative X-component this time, indicating that the pilot needs to set the course slightly west of south to counteract the southwest wind.
Course the pilot must set = arctan(Y-component / X-component) = arctan((-35.355 km/h) / (-250 km/h)) ≈ 8.152° (west of south)
Magnitude of resultant vector = sqrt((X-component)^2 + (Y-component)^2) = sqrt(((-250 km/h)^2) + (-35.355 km/h)^2) ≈ 252.28 km/h
Therefore, on the return journey, the pilot must set a course approximately 8.152° west of south, and the ground speed of the aircraft will be approximately 252.28 km/h.
To find the course the pilot must set and the ground speed of the aircraft, we need to analyze the effect of the wind on the aircraft's motion.
Let's start with the first part of the problem: finding the course the pilot must set and the ground speed on the outbound journey.
1. Course:
To determine the course, we need to account for the wind's direction. The wind blows from the southwest, which is 225 degrees on a compass.
Since the pilot wants to fly due north (360 degrees on a compass), the course he must set will be the angle that compensates for the wind. To find this angle, we can subtract the wind direction from the desired direction as follows:
Course = Desired direction - Wind direction
Course = 360 degrees - 225 degrees
Course = 135 degrees
Therefore, the pilot must set a course of 135 degrees to fly due north.
2. Ground Speed:
The ground speed takes into account the effect of the wind on the aircraft's speed. To find the ground speed, we can use vector addition.
The aircraft's airspeed is given as 250 km/h, and the wind blows from the southwest at 50 km/h. To calculate the ground speed, we need to find the resultant vector of the aircraft's airspeed and the wind velocity.
Using the Pythagorean theorem and trigonometry, we can calculate the ground speed as follows:
Ground Speed = sqrt(airspeed^2 + wind velocity^2 + 2 * airspeed * wind velocity * cos(angle between them))
Plugging in the given values:
Ground Speed = sqrt((250 km/h)^2 + (50 km/h)^2 + 2 * 250 km/h * 50 km/h * cos(135 degrees))
Ground Speed ≈ 330.76 km/h
Therefore, the ground speed of the aircraft on the outbound journey is approximately 330.76 km/h.
Now let's move to the second part of the problem: finding the new course and ground speed on the return journey.
Since the wind speed and direction are unaltered, the wind is still blowing from the southwest at 50 km/h. However, the aircraft will now be flying in the opposite direction (due south).
1. Course:
In this case, to find the new course, we need to account for both the original wind direction and the desired direction of travel.
The original wind direction is southwest (225 degrees), and the desired direction of travel is due south (180 degrees).
To find the new course, we subtract the wind direction from the desired direction:
Course = Desired direction - Wind direction
Course = 180 degrees - 225 degrees
Course = -45 degrees
Since a course cannot be negative, we can also express it as 360 degrees - 45 degrees:
Course = 360 degrees - 45 degrees
Course = 315 degrees
Therefore, the pilot must set a course of 315 degrees to fly due south on the return journey.
2. Ground Speed:
The wind speed and direction remain the same as the outbound journey. Therefore, the ground speed on the return journey will also be the same as before, which is approximately 330.76 km/h.
In summary, the pilot must set a course of 135 degrees to fly due north on the outbound journey, with a ground speed of approximately 330.76 km/h. On the return journey, the pilot must set a course of 315 degrees to fly due south, with the same ground speed of approximately 330.76 km/h.