The car makes a displacement of 100m towards east and then 200m towards north. Find the magnitude and direction of the resultant.

Options

A. 223.7m, `tan^(-1)(2),N` of E
B. 223.7m, `tan^(-1)(2),E` of N
C. 300m, `tan^(-1)(2)`, N of E
D. 100m, `tan^(-1)(2)`, N of E

Explanation

To solve the problem of finding the magnitude and direction of the resultant displacement of the car, we can follow these steps:

### Step 1: Identify the Displacements
The car makes two displacements:
- 100 meters towards the east
- 200 meters towards the north

### Step 2: Represent the Displacements as Vectors
We can represent these displacements as vectors:
- The displacement towards the east can be represented as \( \vec{OA} = 100 \hat{i} \) meters (where \( \hat{i} \) is the unit vector in the east direction).
- The displacement towards the north can be represented as \( \vec{AB} = 200 \hat{j} \) meters (where \( \hat{j} \) is the unit vector in the north direction).

### Step 3: Calculate the Resultant Displacement
The resultant displacement \( \vec{OB} \) can be found by adding the two vectors:
\[ \vec{OB} = \vec{OA} + \vec{AB} = 100 \hat{i} + 200 \hat{j} \]

### Step 4: Calculate the Magnitude of the Resultant Displacement
To find the magnitude of the resultant vector \( \vec{OB} \), we use the Pythagorean theorem:
\[ |\vec{OB}| = \sqrt{(100)^2 + (200)^2} \]
Calculating this gives:
\[ |\vec{OB}| = \sqrt{10000 + 40000} = \sqrt{50000} = 100\sqrt{5} \approx 223.7 \text{ meters} \]

### Step 5: Calculate the Direction of the Resultant Displacement
To find the direction, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{200}{100} = 2 \]
Thus, we find the angle \( \theta \):
\[ \theta = \tan^{-1}(2) \]
This angle is measured north of east.

### Final Result
The magnitude of the resultant displacement is approximately \( 223.7 \) meters, and the direction is \( \tan^{-1}(2) \) north of east.

### Summary
- **Magnitude**: \( 223.7 \) meters
- **Direction**: \( \tan^{-1}(2) \) north of east