a plane with an airspeed of 192mi/h is headed on a bearing of 121 degrees. a north wind is blowing (from north to south) at 15.9 mi/h . find the ground speed and the actual bearing of plane.
This is a vector addition problem. The ground velocity vector (which tells you both ground speed and bearing) is the sum of the air velocity vector and the wind vector (air with respect to ground).
The easiest way to to this kind of problem is adding components; however, solving for the angles of the vector triangle using trigonometry laws can also be used.
I believe that Reiny may gave already solved this problem for you elsewhere.
If not, show your work and someone will check it.
To find the ground speed and actual bearing of the plane, we can use the concept of vector addition.
1. Start by understanding the given information:
- Airspeed: 192 mi/h (speed of the plane with respect to the air)
- Bearing: 121 degrees (direction in which the plane is heading)
- North wind: 15.9 mi/h (speed of the wind blowing from north to south)
2. Break down the airspeed vector into its components:
- Horizontal component: Airspeed * cos(Bearing)
Ground speed in the eastward direction = 192 mi/h * cos(121°)
- Vertical component: Airspeed * sin(Bearing)
Ground speed in the northward direction = 192 mi/h * sin(121°)
3. Add the wind vector to the ground speed components:
- Eastward ground speed = Ground speed in the eastward direction + Wind speed in the eastward direction
- Northward ground speed = Ground speed in the northward direction + Wind speed in the northward direction
4. Calculate the magnitudes of the ground speed:
- Ground speed = Square root of (Eastward ground speed^2 + Northward ground speed^2)
5. Calculate the actual bearing of the plane:
- Actual bearing = arctan(Northward ground speed / Eastward ground speed)
Now, let's calculate the values:
Horizontal component:
Ground speed in the eastward direction = 192 mi/h * cos(121°)
Ground speed in the eastward direction ≈ 192 mi/h * (-0.4848)
Ground speed in the eastward direction ≈ -93.2 mi/h
Vertical component:
Ground speed in the northward direction = 192 mi/h * sin(121°)
Ground speed in the northward direction ≈ 192 mi/h * (0.8746)
Ground speed in the northward direction ≈ 167.8 mi/h
Adding wind vector to ground speed components:
Eastward ground speed = -93.2 mi/h + 0 mi/h (since wind blows north-south)
Northward ground speed = 167.8 mi/h - 15.9 mi/h (subtracting wind speed from northward ground speed)
Calculating the magnitude of the ground speed:
Ground speed = Square root((-93.2 mi/h)^2 + (167.8 mi/h - 15.9 mi/h)^2)
Ground speed ≈ Square root(8657.24 + 2645.44)
Ground speed ≈ Square root(11302.68)
Ground speed ≈ 106.37 mi/h
Calculating the actual bearing of the plane:
Actual bearing ≈ arctan((167.8 mi/h - 15.9 mi/h) / -93.2 mi/h)
Actual bearing ≈ arctan(151.9 mi/h / -93.2 mi/h)
Actual bearing ≈ arctan(-1.63)
Since the tangent function is negative, the actual bearing is in the third or fourth quadrants.
Actual bearing ≈ 189.11°
Therefore, the ground speed of the plane is approximately 106.37 mi/h, and the actual bearing is approximately 189.11 degrees.