An airplane moves at 300 mi/hr relative to the air. It must fly from city A to city B, 600 miles due east. One day the wind is blowing at 100 mi/hr in a direction of 30 degrees north of east. Draw a diagram.

b. What will the velocity of the plane relative to the ground (value and direction)?
c.After the 2 hours flight, how far is city B from the plane, and in what direction?

To answer these questions, we need to break down the velocities and use vector addition. Let's start by drawing a diagram:

1. Draw a line segment to represent the distance from city A to city B, going due east.
2. At city A, draw an arrow pointing east to represent the velocity of the airplane relative to the air, which is 300 mi/hr.
3. Draw another arrow starting at city A, pointing in the direction of the wind, at a 30 degree angle north of east. Label it with a magnitude of 100 mi/hr.
4. Connect the tail of the arrow representing the wind to the head of the arrow representing the velocity of the airplane. This line segment represents the velocity of the airplane relative to the ground.

Now, let's calculate the velocity of the plane relative to the ground:

To find the velocity of the airplane relative to the ground, we need to use vector addition. We can use the cosine and sine laws to decompose the vectors into their east and north components.

Let's call the east component of the velocity of the airplane "Vax" and the north component "Vay." Similarly, let's call the east component of the wind velocity "Vwx" and the north component "Vwy."

We can calculate these components using trigonometry:

Vax = 300 mi/hr * cos(0°) = 300 mi/hr
Vay = 300 mi/hr * sin(0°) = 0 mi/hr
Vwx = 100 mi/hr * cos(30°) = 100 mi/hr * 0.866 = 86.6 mi/hr
Vwy = 100 mi/hr * sin(30°) = 100 mi/hr * 0.5 = 50 mi/hr

Now, we can add these components to find the velocity of the airplane relative to the ground:

Vgx = Vax + Vwx = 300 mi/hr + 86.6 mi/hr = 386.6 mi/hr
Vgy = Vay + Vwy = 0 mi/hr + 50 mi/hr = 50 mi/hr

So, the velocity of the airplane relative to the ground is 386.6 mi/hr to the east and 50 mi/hr north.

To answer part c of the question, we can calculate the displacement of the plane after 2 hours of flight. The displacement is the change in position between the starting point and the final location in terms of distance and direction.

The distance traveled is the magnitude of the velocity vector multiplied by the time:

Distance = Velocity * Time = 386.6 mi/hr * 2 hrs = 773.2 miles

The direction of the displacement can be found using trigonometry. We can use the tangent function to find the angle between the east direction and the direction of the displacement:

tan(θ) = Vgy / Vgx
θ = atan(Vgy / Vgx) = atan(50 mi/hr / 386.6 mi/hr) ≈ 7.4°

Therefore, after 2 hours of flight, city B is approximately 773.2 miles away from the plane in a direction of approximately 7.4° north of east.