A flag is mounted on a car moving due North with velocity of `20 km//hr`. Strong winds are blowing due East with velocity of `20km//gr`. The flag will point in direction

Options

A. East
B. North-East
C. South-East
D. South-West

Answer

To solve the problem of determining the direction in which the flag will point, we can follow these steps:

### Step 1: Understand the velocities involved
- The car is moving due North with a velocity of \(20 \text{ km/hr}\).
- The wind is blowing due East with a velocity of \(20 \text{ km/hr}\).

### Step 2: Represent the velocities as vectors
- The velocity of the car can be represented as a vector pointing upwards (North) on a coordinate system:
\[ \vec{V}_{\text{car}} = (0, 20) \text{ km/hr} \]
- The velocity of the wind can be represented as a vector pointing to the right (East):
\[ \vec{V}_{\text{wind}} = (20, 0) \text{ km/hr} \]

### Step 3: Calculate the resultant velocity vector
- The resultant velocity vector \(\vec{V}_{\text{resultant}}\) can be found by adding the two vectors:
\[ \vec{V}_{\text{resultant}} = \vec{V}_{\text{car}} + \vec{V}_{\text{wind}} = (0 + 20, 20 + 0) = (20, 20) \text{ km/hr} \]

### Step 4: Determine the direction of the resultant vector
- The resultant vector \((20, 20)\) indicates that the flag will point in a direction that is at an angle to both the North and East directions.
- To find the angle \(\theta\) with respect to the North direction, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{20}{20} = 1 \]
- Thus, \(\theta = 45^\circ\).

### Step 5: Identify the direction of the flag
- Since the angle is \(45^\circ\) from the North towards the East, the flag will point in the North-East direction.

### Conclusion
- The flag will point in the North-East direction.