An aeroplane takes off from Mumbai to Delhi with velocity 50 kph in north-east direction. Wind is blowing at 25 kph from north to south. What is the resultant displacement of aeroplane in 2 hrs.

Answer

To solve the problem of the aeroplane's resultant displacement after 2 hours, we will follow these steps:

### Step 1: Understand the Directions and Velocities
The aeroplane is flying in the northeast direction with a velocity of 50 km/h. The wind is blowing from north to south at a velocity of 25 km/h.

### Step 2: Break Down the Velocities into Components
Since the aeroplane is moving in the northeast direction, we can break its velocity into its north and east components using trigonometry.

- The northeast direction makes a 45-degree angle with both the north and east axes.
- The components of the aeroplane's velocity (V1) can be calculated as:
- \( V_{1x} = V_1 \cos(45^\circ) = 50 \times \frac{1}{\sqrt{2}} = \frac{50}{\sqrt{2}} \) km/h (east component)
- \( V_{1y} = V_1 \sin(45^\circ) = 50 \times \frac{1}{\sqrt{2}} = \frac{50}{\sqrt{2}} \) km/h (north component)

### Step 3: Calculate the Effective North Component
The wind is blowing from north to south, which affects the northward component of the aeroplane's velocity. Therefore, we need to subtract the wind's velocity from the north component of the aeroplane's velocity.

- The effective northward velocity (V_eff) is:
- \( V_{eff} = V_{1y} - V_{wind} = \frac{50}{\sqrt{2}} - 25 \) km/h

### Step 4: Calculate the Resultant Velocity
Now we have the eastward component (V1x) and the effective northward component (V_eff). We can find the resultant velocity (V_r) using the Pythagorean theorem:

- \( V_{r} = \sqrt{(V_{1x})^2 + (V_{eff})^2} \)

### Step 5: Calculate the Resultant Displacement in 2 Hours
To find the displacement after 2 hours, we multiply the resultant velocity by the time:

- Displacement = \( V_r \times t \)

### Step 6: Substitute and Calculate
Now we can substitute the values and calculate the resultant displacement.

1. Calculate \( V_{1x} \):
- \( V_{1x} = \frac{50}{\sqrt{2}} \approx 35.36 \) km/h

2. Calculate \( V_{eff} \):
- \( V_{eff} = 35.36 - 25 = 10.36 \) km/h

3. Calculate \( V_{r} \):
- \( V_{r} = \sqrt{(35.36)^2 + (10.36)^2} \)
- \( V_{r} = \sqrt{1252.57 + 107.36} = \sqrt{1359.93} \approx 36.84 \) km/h

4. Calculate the displacement in 2 hours:
- Displacement = \( 36.84 \times 2 = 73.68 \) km

### Final Result
The resultant displacement of the aeroplane in 2 hours is approximately **73.68 km**.

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