a jet airliner moving initially at 3..00X10^2 mi/h due east enters a region where the wind is blowing 1.0 x 10^2 mi/h in a direction 30 degrees north of east. A) find the components of the velocity of the jet airliner relative to the air. b) find the components of the velocity of the air relative to the earth.C) what are the speed and direction of the aircraft relative to the ground
To find the components of the velocity of the jet airliner relative to the air, we need to break down the velocity into its x and y components.
Let's assume the x-axis is due east and the y-axis is due north. The initial velocity of the jet airliner is 3.00x10^2 mi/h due east. Since there is no wind at this point, the x-component of the velocity relative to the air is 3.00x10^2 mi/h.
Now, let's consider the wind. The wind is blowing at 1.0x10^2 mi/h in a direction 30 degrees north of east. To find the y-component of the velocity relative to the air, we can use trigonometry.
The y-component of the velocity can be found using the formula: velocity * sin(angle).
Here, the velocity is 1.0x10^2 mi/h and the angle is 30 degrees.
So, the y-component of the velocity relative to the air is (1.0x10^2 mi/h) * sin(30°) = 50 mi/h.
Therefore, the components of the velocity of the jet airliner relative to the air are:
- x-component: 3.00x10^2 mi/h due east.
- y-component: 50 mi/h due north.
To find the components of the velocity of the air relative to the earth, we need to subtract the velocity of the wind from the components of the velocity of the jet airliner relative to the air.
- x-component of the velocity of the air relative to the earth: 3.00x10^2 mi/h - 0 (since the wind is blowing in the y-direction).
- y-component of the velocity of the air relative to the earth: 50 mi/h - 1.0x10^2 mi/h (opposite direction of the wind).
Thus, the components of the velocity of the air relative to the earth are:
- x-component: 3.00x10^2 mi/h.
- y-component: -50 mi/h (due south since it is opposite to the wind direction).
To find the speed and direction of the aircraft relative to the ground, we can use the Pythagorean theorem and trigonometry.
The speed (magnitude of velocity) relative to the ground can be found using the formula: speed = sqrt(x^2 + y^2).
Here, x-component = 3.00x10^2 mi/h and y-component = -50 mi/h.
So, the speed of the aircraft relative to the ground is sqrt((3.00x10^2)^2 + (-50)^2).
And the direction of the aircraft relative to the ground can be found using the formula: angle = atan(y/x).
Here, x-component = 3.00x10^2 mi/h and y-component = -50 mi/h.
So, the direction of the aircraft relative to the ground is atan((-50)/(3.00x10^2)).
After calculating these values, you will have the speed and direction of the aircraft relative to the ground.