A plane flies due north (90° from east) with a velocity of 100 km/h for 3 hours. During this time, a steady wind blows southeast at 30 km/h at an angle of 315° from due east. After 3 hours, where will the plane’s position be relative to its starting point? Show your work.
v=sqrt{v(p)²+v(w)²-2v(p)v(w)cos α}=
=sqrt{100²+30²-2•100•30cos45} =81.6 km/h
s= vt=81.6•3=244.8 km
β= 85°v (northeast)
Well, let's break it down, my friend. The plane flies due north for 3 hours at a velocity of 100 km/h. So, in those 3 hours, it covers a distance of 300 km.
Now, let's talk about the wind. It blows southeast at 30 km/h at an angle of 315° from due east. That means it's pushing the plane a bit to the southwest.
To find out where the plane will be after 3 hours, we need to consider both the plane's velocity and the wind's effect. We can break down the plane's velocity into two components: one in the north direction and one in the east direction.
The north component would be 100 km/h * cos(90°) = 0 km/h, because its velocity is purely in the north direction. The east component would be 100 km/h * sin(90°) = 100 km/h, because it has no velocity in the east direction.
Now, let's look at the wind's effect. It is blowing at an angle of 315° from due east, so its east component would be 30 km/h * cos(45°) = 21.2 km/h, and its north component would be 30 km/h * sin(45°) = 21.2 km/h.
To find the resultant velocity of the plane, we add the respective components of the plane and the wind together. The new east component would be 100 km/h + 21.2 km/h = 121.2 km/h, and the new north component would be 0 km/h + 21.2 km/h = 21.2 km/h.
After 3 hours, the plane would have traveled 121.2 km/h * 3 hours = 363.6 km eastward and 21.2 km/h * 3 hours = 63.6 km northward.
So, the plane's position would be 363.6 km eastward and 63.6 km northward relative to its starting point. But hey, don't worry about getting lost with all these numbers. As a clown, I'm always here to navigate you with humor!
To find the plane's position after 3 hours, we need to analyze the velocity components.
1. Plane's velocity:
The plane flies due north with a velocity of 100 km/h. Since it's only moving in the north direction, the eastward component of its velocity is 0 km/h.
2. Wind's velocity:
The wind blows southeast at an angle of 315° from due east, with a magnitude of 30 km/h. To find the velocity components in the eastward (x) and northward (y) directions, we can use trigonometry.
- The eastward component: 30 km/h * cos(45°) = 21.21 km/h
- The northward component: 30 km/h * sin(45°) = 21.21 km/h
3. Resultant velocity:
To find the resultant velocity, we can add the velocity components of the plane and the wind:
- The eastward component: 0 km/h + 21.21 km/h = 21.21 km/h
- The northward component: 100 km/h + 21.21 km/h = 121.21 km/h
The resultant velocity is √(21.21^2 + 121.21^2) ≈ 123.78 km/h.
4. Displacement:
The displacement can be found by multiplying the resultant velocity by the time, which is 3 hours:
- The eastward displacement: 21.21 km/h * 3 h = 63.63 km
- The northward displacement: 121.21 km/h * 3 h = 363.63 km
5. Final position:
To determine the final position relative to the starting point, we put the eastward displacement as the x-coordinate and the northward displacement as the y-coordinate. So, after 3 hours, the plane's position will be approximately at (63.63 km, 363.63 km) relative to its starting point.
Note: The angle of 315° is equivalent to π/4 radians or 45°.
To determine the plane's position after 3 hours, we need to consider both the velocity of the plane and the velocity of the wind.
The velocity of the plane can be represented as a vector pointing due north with a magnitude of 100 km/h. The velocity of the wind can be represented as a vector pointing southeast with a magnitude of 30 km/h.
We can break down the velocity of the wind into its north and east components using trigonometry. The angle of 315° from due east is equivalent to an angle of 45° from due north.
The north component of the wind's velocity is given by:
North component = Wind velocity * sin(angle)
North component = 30 km/h * sin(45°)
North component = 30 km/h * √(2)/2
North component ≈ 21.21 km/h (rounded to the nearest hundredth)
The east component of the wind's velocity is given by:
East component = Wind velocity * cos(angle)
East component = 30 km/h * cos(45°)
East component = 30 km/h * √(2)/2
East component ≈ 21.21 km/h (rounded to the nearest hundredth)
Now, we can add the velocities of the plane and the wind to find the resultant velocity.
The north component of the resultant velocity is given by:
North component = Plane velocity + North component of wind velocity
North component = 100 km/h + 21.21 km/h
North component = 121.21 km/h
The east component of the resultant velocity is given by:
East component = East component of wind velocity
East component = 21.21 km/h
Therefore, after 3 hours, the plane's position will be 121.21 km north and 21.21 km east from its starting point, relative to a due north-east reference point.