A pilot must fly his plane to a town which is 200 km from his starting point in a direction 30 degrees N of E. He must make the trip in 1/2 hr. An 80km/h wind blows in a direction 30 degrees E of S. Find the speed of the plane relative to the ground, the speed relative to air and the heading of the plane.
The answers are 400 km/h, 408 km/h and 41 degrees N of E. I don't know how to do this problem. I think I did my diagram wrong.
One method
Three velocity vectors: the wind, the plane, and the desired results. Since the plane must reach the town in 1/2 hour, the results must be 400km/hr at 30 degrees N of E. That defines the speed over the ground.
Draw the results vector.
Now draw the wind vector as given. Handily for this problem the wind and the results are at 90 degrees. Draw the planes vector from the end of the wind vector to the end of the results vector. The length is the plane's speed. The heading of the plane is the angle where the plane's vector crosses the x-axis. There are several ways to find this angle. Be careful. It is not equal to any of the angles in the triangle just solved, but you do need them to find it.
The heading solved will be a heading relative to east (chosen since the original problem used that). You can adjust that to a heading with 0 degrees at true north if desired.
To solve this problem, we can break it down into two components: the horizontal motion and the vertical motion. Let's tackle the horizontal motion first.
1. Horizontal Motion:
The pilot wants to fly in a direction 30 degrees North of East. We can represent this direction as an angle with respect to the positive x-axis, which is the East direction. So, the angle of the plane's heading is 30 degrees.
Let's denote the speed of the plane relative to the ground as Vp, and the speed of the wind as Vw.
The horizontal component of the plane's speed relative to the ground will be given by:
Vpx = Vp * cos(30 degrees)
The horizontal component of the wind's speed will be given by:
Vwx = -Vw * sin(30 degrees)
The negative sign (-) is used because the wind is blowing in the opposite direction, which is 30 degrees East of South, so we consider it in the negative x-axis direction.
Now, we can find the horizontal speed of the plane relative to the ground:
Vx = Vpx + Vwx
2. Vertical Motion:
Since we are given the time it takes for the pilot to travel, we can calculate the distance the plane covers vertically.
The vertical component of the plane's speed relative to the ground will be given by:
Vpy = Vp * sin(30 degrees)
Now, we can find the vertical distance traveled by the plane:
Vy = Vpy * time
To find the total distance traveled by the plane, we can use Pythagoras' theorem:
Total distance = √(Vx² + Vy²)
Now, let's calculate the values step by step:
1. Calculate the horizontal component of the plane's speed relative to the ground:
Vpx = Vp * cos(30 degrees)
2. Calculate the horizontal component of the wind's speed:
Vwx = -Vw * sin(30 degrees)
3. Calculate the horizontal speed of the plane relative to the ground:
Vx = Vpx + Vwx
4. Calculate the vertical component of the plane's speed relative to the ground:
Vpy = Vp * sin(30 degrees)
5. Calculate the vertical distance traveled by the plane:
Vy = Vpy * time (time given as 1/2 hour)
6. Calculate the total distance traveled by the plane:
Total distance = √(Vx² + Vy²)
Now, we can substitute the given values into these equations and solve for the unknowns to get the speed of the plane relative to the ground, the speed of the plane relative to the air, and the heading of the plane.