An airplane is heading due east. The airspeed indicator shows that the plane is moving at a speed of 370 km/h relative to the air. If the wind is blowing from the north at 92.5 km/h, the velocity of the airplane relative to the ground is
370 E and 92.5 S
|v| = sqrt (370^2 + 92.5^2)
tan angle south of east = 92.5/370 = 0.25
angle south of east = 14 deg
angle clockwise from north = 90 + 14 = 104 deg True
To determine the velocity of the airplane relative to the ground, we need to consider the effect of the wind. Since the airplane is moving due east and the wind is blowing from the north, these two velocities will combine to give us the resultant velocity.
We can use vector addition to find the resultant velocity:
1. First, let's break down the velocities into their x and y components:
- The airspeed of the plane is 370 km/h due east, so its x-component is 370 km/h and its y-component is 0 km/h.
- The wind is blowing from the north at 92.5 km/h, so its x-component is 0 km/h and its y-component is 92.5 km/h.
2. Next, we add the x-components and y-components separately:
- X-component: 370 km/h + 0 km/h = 370 km/h (still due east)
- Y-component: 0 km/h + 92.5 km/h = 92.5 km/h (northward)
3. Finally, we can combine the x-component and y-component to find the resultant velocity:
- The resultant velocity is the square root of the sum of the squares of the x and y components:
Resultant velocity = sqrt((370 km/h)^2 + (92.5 km/h)^2)
= sqrt(136900 + 8556.25)
= sqrt(145456.25)
≈ 381.56 km/h
Therefore, the velocity of the airplane relative to the ground is approximately 381.56 km/h. It has a direction of around 7.19° north of east.
To find the velocity of the airplane relative to the ground, we need to use vector addition.
First, let's break down the velocities into their respective components:
- The eastward velocity of the airplane relative to the air is 370 km/h.
- The northward velocity of the wind is 92.5 km/h.
Now, let's construct a right triangle to visualize the velocities. The eastward velocity of the airplane is one side of the triangle, the northward velocity of the wind is the other side, and the resulting velocity (relative to the ground) is the hypotenuse. Since the airplane is heading due east, the triangle will have a right angle.
Using the Pythagorean theorem, we can calculate the magnitude of the resulting velocity:
Resulting velocity = √(eastward velocity^2 + northward velocity^2)
Resulting velocity = √(370^2 + 92.5^2)
Resulting velocity ≈ √(136900 + 8556.25)
Resulting velocity ≈ √145456.25
Resulting velocity ≈ 381.25 km/h
Therefore, the velocity of the airplane relative to the ground is approximately 381.25 km/h.