An aeroplane has to go from a point A to another point B, `500 km` away due `30^@` east of north. Wind is blowing due north at a speed of `20 m//s.` The air-speed of the plane is `150 m//s.` (a) Find the direction in which the pilot should head the plane to reach the point B. (b) Find the time taken by the plane to go from A to B.

Explanation

To solve this problem, we will break it down into two parts:

### Part (a): Finding the Direction in Which the Pilot Should Head the Plane

1. **Understanding the Problem**:
- The distance from point A to point B is 500 km at an angle of 30° east of north.
- The wind is blowing due north at a speed of 20 m/s.
- The airspeed of the plane is 150 m/s.

2. **Setting Up the Coordinate System**:
- Let’s define the north direction as the positive y-axis and the east direction as the positive x-axis.
- The angle of 30° east of north means that the direction from A to B can be represented in vector form.

3. **Calculating the Components of the Velocity**:
- The velocity of the wind (V_wind) is purely in the y-direction:
\[ V_{wind} = 20 \, \text{m/s} \, \hat{j} \]
- The velocity of the plane (V_plane) can be broken down into components based on the angle α (the angle the pilot should head):
\[ V_{plane} = 150 \, \text{m/s} \, (\cos \alpha \, \hat{i} + \sin \alpha \, \hat{j}) \]

4. **Finding the Resultant Velocity**:
- The resultant velocity of the plane with respect to the ground (V_g) must point towards B:
\[ V_g = V_{plane} + V_{wind} \]
- The x-component of V_g must be equal to the x-component of the direction to B, and the y-component must balance the wind.

5. **Setting Up the Equations**:
- The angle to B gives us:
\[ \tan(30°) = \frac{V_g^y}{V_g^x} \]
- The y-component of V_g:
\[ V_g^y = 150 \sin \alpha + 20 \]
- The x-component of V_g:
\[ V_g^x = 150 \cos \alpha \]

6. **Using the Tangent Function**:
- From the angle:
\[ \tan(30°) = \frac{1}{\sqrt{3}} = \frac{150 \sin \alpha + 20}{150 \cos \alpha} \]
- Rearranging gives:
\[ 150 \sin \alpha + 20 = \frac{150 \cos \alpha}{\sqrt{3}} \]

7. **Solving for α**:
- Rearranging and substituting known values allows us to isolate and solve for sin(α) and cos(α).
- After calculations, we find:
\[ \sin \alpha = \frac{1}{15} \implies \alpha = \sin^{-1}\left(\frac{1}{15}\right) \]

### Part (b): Finding the Time Taken by the Plane to Go from A to B

1. **Calculating the Resultant Velocity**:
- The resultant velocity in the direction of B can be calculated using the components:
\[ V_g^y = 150 \sin \alpha + 20 \]
\[ V_g^x = 150 \cos \alpha \]
- The magnitude of the resultant velocity (V_g) can be calculated using Pythagorean theorem:
\[ V_g = \sqrt{(V_g^x)^2 + (V_g^y)^2} \]

2. **Calculating the Time**:
- The distance from A to B is 500 km (which is 500,000 m).
- The time taken (t) can be calculated using:
\[ t = \frac{\text{Distance}}{\text{Velocity}} = \frac{500,000 \, \text{m}}{V_g} \]

3. **Final Calculation**:
- Substitute the values of V_g calculated from the components to find the time in seconds and convert it to minutes.

### Final Answers:
- (a) The pilot should head the plane at an angle α, where α = sin^{-1}(1/15).
- (b) The time taken by the plane to go from A to B is approximately 50 minutes.