An aeroplane takes off from Mumbai to Delhi with velocity 50 kph at an angle `30^(@)` north of east. Wind is blowing at 25 kph from north to south. What is the resultant displacement of aeroplane in 2 hrs. (in km)

Explanation

To solve the problem step by step, we will break down the components of the airplane's velocity and the effect of the wind, and then calculate the resultant displacement over a period of 2 hours.

### Step 1: Determine the components of the airplane's velocity
The airplane's velocity is given as 50 km/h at an angle of 30 degrees north of east. We can resolve this velocity into its eastward (x-direction) and northward (y-direction) components using trigonometric functions.

- **Eastward component (Vx)**:
\[ V_x = V \cdot \cos(\theta) = 50 \cdot \cos(30^\circ) \]
\[ V_x = 50 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3} \text{ km/h} \]

- **Northward component (Vy)**:
\[ V_y = V \cdot \sin(\theta) = 50 \cdot \sin(30^\circ) \]
\[ V_y = 50 \cdot \frac{1}{2} = 25 \text{ km/h} \]

### Step 2: Account for the wind's effect
The wind is blowing from north to south at a speed of 25 km/h. This will affect the northward component of the airplane's velocity.

- The effective northward velocity (Vy') of the airplane is:
\[ V_y' = V_y - \text{wind speed} = 25 - 25 = 0 \text{ km/h} \]

### Step 3: Calculate the resultant velocity of the airplane
Now we can find the resultant velocity vector of the airplane. Since the northward component is 0, the resultant velocity is purely in the eastward direction.

- **Resultant velocity (V)**:
\[ V = V_x + V_y' = 25\sqrt{3} \text{ km/h} + 0 = 25\sqrt{3} \text{ km/h} \]

### Step 4: Calculate the displacement over 2 hours
Displacement can be calculated using the formula:
\[ \text{Displacement} = \text{Velocity} \times \text{Time} \]
Given that the time is 2 hours:
\[ \text{Displacement} = 25\sqrt{3} \times 2 \text{ km} \]
\[ \text{Displacement} = 50\sqrt{3} \text{ km} \]

### Step 5: Calculate the numerical value of the displacement
Using the approximate value of \(\sqrt{3} \approx 1.73\):
\[ \text{Displacement} \approx 50 \times 1.73 = 86.5 \text{ km} \]

### Conclusion
The resultant displacement of the airplane in 2 hours is approximately **86.5 km** towards the east.

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