An airplane has an airspeed of 770 kilometers per hour at a bearing of N 48 degrees E. If the wind velocity is 28 kilometers per hour from the west, find the angle representing the bearing for
the ground speed?
I found Rx and Ry which are 543.23 and 572.22 respectively. To find the angle I took tan^-1 of 572.22/543.23 and got 46. 489 for the angle, but the answer is wrong. R=789 and I got that right so I don't understand why the angle is incorrect.
Vp + 28[0o] = 770[90-48] = 770[42o]CCW
Vp + 28 = 770*cos42 + i770*sin42
Vp + 28 = 572.2 + 515.2i
Vp = 544.2 + 515.2i = 749mi/h[43.6o] CCW
= 749mi/h[46.4o] E of N.
To find the angle representing the bearing for the ground speed in this scenario, we need to consider the vector addition of the airplane's airspeed and the wind velocity.
1. Start by drawing a diagram to visualize the scenario. Place the airplane's airspeed vector, pointing in the direction of N 48 degrees E, and label it as A. Also, draw the wind velocity vector, pointing towards the west, and label it as W.
2. The ground speed vector, which represents the resultant of the airspeed and the wind velocity, is the vector sum of A and W. Let's label it as G.
3. Use vector addition to find the magnitude and direction of the ground speed. To perform vector addition, we can break down the vectors into their x-components and y-components.
The airspeed vector, A, has an airspeed of 770 kilometers per hour at a bearing of N 48 degrees E. To find the x-component and y-component of A, we can use trigonometric functions.
- The x-component of A (Ax) can be found using the formula Ax = |A| * cos(theta), where |A| represents the magnitude of A and theta represents the angle N 48 degrees E.
- The y-component of A (Ay) can be found using the formula Ay = |A| * sin(theta).
Given that |A| = 770 km/h and theta = 48 degrees, we can calculate Ax and Ay as follows:
Ax = 770 km/h * cos(48 degrees)
Ay = 770 km/h * sin(48 degrees)
Calculating these values, we get:
Ax ≈ 543.23 km/h
Ay ≈ 572.22 km/h
Now, let's consider the wind velocity vector, W. It has a velocity of 28 kilometers per hour towards the west, which means its x-component, Wx, will be -28 km/h, and its y-component, Wy, will be 0 km/h.
4. Add the x-components and y-components of A and W to find the x-component and y-component of the ground speed vector, G.
Gx = Ax + Wx
Gy = Ay + Wy
Substituting the values we found earlier, we get:
Gx = 543.23 km/h + (-28 km/h) ≈ 515.23 km/h
Gy = 572.22 km/h + 0 km/h ≈ 572.22 km/h
Therefore, the x-component of the ground speed is approximately 515.23 km/h, and the y-component is approximately 572.22 km/h.
5. To find the magnitude of the ground speed, use the Pythagorean theorem.
The magnitude of the ground speed, |G|, is given by the formula:
|G| = sqrt(Gx^2 + Gy^2)
Substituting the values we found earlier, we have:
|G| = sqrt((515.23 km/h)^2 + (572.22 km/h)^2)
Calculating this, we get:
|G| ≈ 789 km/h
Therefore, the magnitude of the ground speed is approximately 789 km/h.
6. Finally, to find the angle representing the bearing for the ground speed, we can use the inverse tangent function.
The angle, theta, is given by the formula:
theta = tan^-1(Gy / Gx)
Substituting the values we found earlier, we have:
theta = tan^-1(572.22 km/h / 515.23 km/h)
Calculating this, we get:
theta ≈ 48.489 degrees
Therefore, the angle representing the bearing for the ground speed is approximately 48.489 degrees.
It seems like you made a mistake in the calculations. The correct angle representing the bearing for the ground speed should be approximately 48.489 degrees, which is consistent with the initial bearing of N 48 degrees E.