can someone take me through this vector combination question please
an airplane is observed to be flying at a speed of 600km/h. The planes nose points west. The winds velcoity is 40km/h 45degrees west of south. Find the air planes velocity relative to the ground.
velocity with respect to ground =
(plane velocity with respect to air) + (air (wind) velocity with respect to ground).
You need to do the vector addition. Unfortunately the statement "the plane is observed to be flying at a speed of 600 km/h" is ambiguous. They don't say who is doing the "observing", or how. If it is seem from the ground, it is what it is, 600 km/h. That could also be a ground speed determined by an onboard GPS sensor. They should be quoting an air speed, and perhaps that is what they attended. The air speed would be toward the west, where the plane is pointed.
This a poorly posed question. That is the fault of your course provider, not yours.
what do i do...
Assume the "observed velocity" is the "velocity with respect to the air", with magnitude 600 km/h and direction west. Then proceed to do the vector addition suggested.
velocity in air: -600 i
air velocity: -20 sqrt2 i - 20 sqrt2 j
Vector sum: -628.3 i - 28.3 j
"i" is a unit vector west and "j" is a unit vector north
The magnitude of the ground velocity is
sqrt[(628.3)^2 + (28.3)^2]= 629 km/h
The direction is south of west by an angle arctan 28.3/228.3 = 7 degrees
Sure! To find the airplane's velocity relative to the ground, we need to combine its velocity (600 km/h west) with the wind's velocity (40 km/h 45 degrees west of south).
First, let's break down both velocities into their horizontal and vertical components.
The airplane's velocity (600 km/h west) has a horizontal component (-600 km/h) and a vertical component (0 km/h) since it is only moving westward.
The wind's velocity (40 km/h 45 degrees west of south) can be broken down into its horizontal and vertical components using trigonometry. The horizontal component will be 40 km/h * sin(45) = 40 km/h * (√2 / 2) ≈ 28.3 km/h toward the west. The vertical component will be 40 km/h * cos(45) = 40 km/h * (√2 / 2) ≈ 28.3 km/h toward the south.
Now, let's add the horizontal components and the vertical components separately to get the total horizontal and vertical velocity, which will give us the airplane's velocity relative to the ground.
Horizontal component: -600 km/h + (-28.3 km/h) = -628.3 km/h (westward)
Vertical component: 0 km/h + (-28.3 km/h) = -28.3 km/h (southward)
The total velocity of the airplane relative to the ground is the vector sum of these horizontal and vertical components. To find the magnitude and direction of this resulting vector, we can use the Pythagorean theorem and trigonometry.
Magnitude of the resulting vector: √((-628.3 km/h)^2 + (-28.3 km/h)^2) ≈ 631.6 km/h
Direction of the resulting vector: arctan((-28.3 km/h) / (-628.3 km/h)) ≈ 2.37 degrees south of west
Therefore, the airplane's velocity relative to the ground is approximately 631.6 km/h, directed 2.37 degrees south of west.