An airplane with an airspeed 600km/h needs to be on a course 35.0degree north of East. A steady wind of 100km/h is blowing from North. Determine the direction the plan should head.
[Give the answer with a figure and proper steps]
so you want
<600 sinθ,600cosθ> + <0,-100> = <v sin35°,v cos35°>
To determine the direction the plane should head, we need to find the resultant velocity vector of the plane.
Step 1: Start by drawing a diagram to represent the situation. Label the vectors involved.
Let's draw a diagram with a horizontal line representing the east direction, a vertical line representing the north direction, and an angle of 35.0 degrees north of east for the plane's desired course. Label the wind vector as "W" and the plane's airspeed vector as "A."
```
----W---->
|\
| \
| \
A | \
|
|
V
```
Step 2: Break down the vectors into their x and y components.
The wind vector is pointing straight down from the north, so its x-component is 0, and its y-component is -100 km/h.
The airspeed vector is at an angle of 35.0 degrees north of east. To find its x and y components, we use trigonometry.
The x-component (A_x) can be found by multiplying the magnitude of the airspeed (600 km/h) by the cosine of the angle (35.0 degrees):
A_x = 600 km/h * cos(35.0 degrees)
The y-component (A_y) can be found by multiplying the magnitude of the airspeed (600 km/h) by the sine of the angle (35.0 degrees):
A_y = 600 km/h * sin(35.0 degrees)
Step 3: Calculate the resultant vector by adding the x and y components.
To find the resultant vector, we add the x components to get the x-component of the resultant vector, and we add the y components to get the y-component of the resultant vector.
Resultant x-component (R_x) = A_x + 0 (wind has no x-component)
Resultant y-component (R_y) = A_y + (-100 km/h) (negative direction due to wind blowing north)
Step 4: Calculate the magnitude and direction of the resultant vector.
The magnitude of the resultant vector can be found using the Pythagorean theorem:
Resultant magnitude = √(R_x^2 + R_y^2)
The direction of the resultant vector can be found using trigonometry:
Resultant direction = arctan(R_y / R_x)
Substitute the calculated values of R_x and R_y into the above formulas to find the magnitude and direction of the resultant vector.
Step 5: Determine the direction the plane should head.
The direction the plane should head is the resultant direction PLUS 180 degrees, since we want the direction relative to the east.
Let's calculate the values:
A_x = 600 km/h * cos(35.0 degrees) = 491.42 km/h
A_y = 600 km/h * sin(35.0 degrees) = 342.51 km/h
R_x = A_x + 0 = 491.42 km/h
R_y = A_y + (-100 km/h) = 242.51 km/h
Resultant magnitude = √(R_x^2 + R_y^2) = √(491.42^2 + 242.51^2) = 542.02 km/h
Resultant direction = arctan(R_y / R_x) = arctan(242.51 / 491.42) = 26.30 degrees
Direction plane should head = Resultant direction + 180 degrees = 26.30 degrees + 180 degrees = 206.30 degrees
Therefore, the plane should head in a direction 206.30 degrees north of east.