An airplane pilot wishes to fly due west. A wind of 80km/hr is blowing towards the south.

a) If the speed of llane in air is 320km/hr, in which direction should the pilot head?

b) What is the speed of plane over the ground?

(Illustrate with diagram)

To solve this problem, we can break it down into two components: the motion of the airplane in the air and the effect of the wind on the airplane.

a) To determine the direction in which the pilot should head, we need to find the resultant velocity vector by combining the velocity of the airplane in the air with the velocity of the wind. Since the pilot wishes to fly due west, we need to find the wind component that cancels out the north/south component of the airplane's velocity.

We can represent the velocity of the airplane as a vector pointing due west with a magnitude of 320 km/hr. Since the wind is blowing towards the south, we can represent the velocity of the wind as a vector pointing due south with a magnitude of 80 km/hr.

To find the resultant velocity vector, we subtract the wind vector from the velocity of the airplane vector. The equation can be written as follows:

Resultant velocity = Velocity of airplane - Velocity of wind

In this case, the resultant velocity vector will have a component towards the west and a component towards the south.

b) To find the speed of the plane over the ground (groundspeed), we need to calculate the magnitude of the resultant velocity vector. This will give us the speed of the plane considering both its velocity in the air and the effect of the wind.

To illustrate this, you can draw a triangle on a diagram. The base of the triangle represents the velocity of the airplane towards the west (320 km/hr), the height represents the velocity of the wind towards the south (80 km/hr), and the hypotenuse represents the resultant velocity vector.

Using the Pythagorean theorem, we can calculate the magnitude of the resultant velocity vector:

Resultant velocity magnitude = √(Velocity of airplane)^2 + (Velocity of wind)^2

Once you have calculated the resultant velocity magnitude, you will have the speed of the plane over the ground.

Note: It's important to keep in mind that the direction in which the pilot should head (question a) will be the opposite direction to the resultant velocity vector, while the speed of the plane over the ground (question b) corresponds to the magnitude of the resultant velocity vector.

To find the direction in which the pilot should head, we need to consider the resultant of the plane's velocity and the wind's velocity.

a) The speed of the plane in the air is 320 km/hr to the west. The wind is blowing towards the south with a speed of 80 km/hr.

To find the resultant, we can use vector addition. Draw a diagram representing the situation, considering the wind blowing towards the south:

```
|
|
| Wind
|
-----PLANE--------> (Towards the west)
```

Now, the wind vector is pointing towards the south (downwards) and has a velocity of 80 km/hr. The plane vector is towards the west and has a velocity of 320 km/hr.

Using vector addition, draw the resultant vector by connecting the initial point of the plane vector to the final point of the wind vector. The direction of this resultant vector represents the direction in which the pilot should head.

```
|
|
| Wind
|
\ |
\|
\
\
\
- RESULTANT (Direction in which the pilot should head)
```

The resultant vector should be inclined towards south-west direction. This means the pilot should head in the south-west direction.

b) To find the speed of the plane over the ground, we can use the Pythagorean theorem. The speed of the plane over the ground is the magnitude of the resultant vector.

Using the Pythagorean theorem:

(Resultant)^2 = (Velocity of plane in air)^2 + (Velocity of wind)^2

(Resultant)^2 = (320 km/hr)^2 + (80 km/hr)^2

(Resultant)^2 = 102,400 km^2/hr^2 + 6,400 km^2/hr^2

(Resultant)^2 = 108,800 km^2/hr^2

(Resultant) = sqrt(108,800) km/hr

(Resultant) ≈ 329.73 km/hr

So, the speed of the plane over the ground is approximately 329.73 km/hr.

draw the diagram.

I see a right triangle.
a. sinTheta=80/320
theta= arcsin(.25)
so direction is theta N of W.

b. speed=sqrt(320^2-80^2)