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

Options

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

Answer

To solve the problem of finding the magnitude and direction of the resultant displacement of a car that moves 100 m east and then 200 m north, we can follow these steps:

### Step 1: Understand the Displacement Vectors
The car first moves 100 m towards the east, which we can denote as vector \( \vec{A} \). Then it moves 200 m towards the north, which we can denote as vector \( \vec{B} \).

### Step 2: Represent the Displacements
- Let \( \vec{A} = 100 \, \text{m} \) (east)
- Let \( \vec{B} = 200 \, \text{m} \) (north)

### Step 3: Use the Pythagorean Theorem to Find the Resultant Magnitude
The resultant displacement \( \vec{R} \) can be found using the Pythagorean theorem since the two displacements are perpendicular to each other:

\[ R = \sqrt{A^2 + B^2} \]

Substituting the values:

\[ R = \sqrt{(100 \, \text{m})^2 + (200 \, \text{m})^2} \]

Calculating the squares:

\[ R = \sqrt{10000 + 40000} \]
\[ R = \sqrt{50000} \]
\[ R = 100 \sqrt{5} \, \text{m} \]

### Step 4: Calculate the Numerical Value of the Resultant
To find the numerical value of \( R \):

\[ R \approx 100 \times 2.236 = 223.6 \, \text{m} \]

### Step 5: Determine the Direction of the Resultant
To find the direction, we can use the tangent function:

\[ \tan(\theta) = \frac{B}{A} = \frac{200}{100} = 2 \]

Now, we find the angle \( \theta \):

\[ \theta = \tan^{-1}(2) \]

Calculating \( \tan^{-1}(2) \):

\[ \theta \approx 63.4^\circ \]

### Step 6: State the Direction
The direction is measured from the east towards the north, so we can say the direction is \( 63.4^\circ \) north of east.

### Final Result
The magnitude of the resultant displacement is approximately \( 223.6 \, \text{m} \) and the direction is \( 63.4^\circ \) north of east.

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