A plane flies on a heading of N25W at a constant speed of 520 km/h. if the velocity of the wind is 60km/h from E30N, what is the velocity of the plane with respect to the ground?
We can solve this problem using vector addition. We start by drawing a diagram:
```
60 km/h
E30N
|
|
25 degrees | 155 degrees
N -------------------- W
|
|
520 km/h
```
We represent the plane's velocity as a vector pointing to the northwest, with a magnitude of 520 km/h. We represent the velocity of the wind as a vector pointing to the southeast, with a magnitude of 60 km/h. We can then add these vectors using vector addition.
To do this, we need to resolve the velocity of the wind into its north-south and east-west components. We can do this using trigonometry:
```
sin(30) = north component / 60
north component = 60 * sin(30) = 30 km/h
cos(30) = east component / 60
east component = 60 * cos(30) = 52.1 km/h
```
So the velocity of the wind is 30 km/h to the north and 52.1 km/h to the east.
We can now add the vectors using the head-to-tail method. We start by placing the tail of the wind vector at the head of the plane vector, and draw the resulting vector from the tail of the plane vector to the head of the wind vector.
```
60 km/h
E30N
|
\--------+
\ |
\ |
\ |
\ |
\ |
\ |
\ |
\|
X
```
The resulting vector represents the velocity of the plane with respect to the ground. To find its magnitude and direction, we can use trigonometry again:
```
tan(25) = north component / west component
north component = west component * tan(25)
magnitude = sqrt(west component^2 + north component^2)
magnitude = sqrt((520 - 52.1)^2 + (52.1 * tan(25))^2) = 478.9 km/h
cos(angle) = west component / magnitude
angle = acos(west component / magnitude)
angle = acos((520 - 52.1) / 478.9) = 61.4 degrees
The velocity of the plane with respect to the ground is 478.9 km/h at an angle of 61.4 degrees west of north.
```