Two cars A and B start moving from the same point with same velocity v=5km/minute. Car A moves towards North and car B in moving towards East. What is the relative velocity of B with respect to A ?

Options

A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-I
B. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-I
C. Statement-I is true, Statement-II is false
D. Statement-I is false, Statement-II is true.

Explanation

To solve the problem of finding the relative velocity of car B with respect to car A, we can follow these steps:

### Step 1: Understand the Directions
Car A is moving towards the North, and Car B is moving towards the East. We can represent these directions using a coordinate system where:
- North corresponds to the positive y-axis.
- East corresponds to the positive x-axis.

### Step 2: Represent the Velocities as Vectors
Given that both cars have the same speed \( v = 5 \) km/min, we can express their velocities as vectors:
- Velocity of Car A, \( \vec{v_A} = 5 \hat{j} \) km/min (since it moves north).
- Velocity of Car B, \( \vec{v_B} = 5 \hat{i} \) km/min (since it moves east).

### Step 3: Calculate the Relative Velocity
The relative velocity of Car B with respect to Car A is given by the formula:
\[ \vec{v_{BA}} = \vec{v_B} - \vec{v_A} \]
Substituting the values:
\[ \vec{v_{BA}} = 5 \hat{i} - 5 \hat{j} \]

### Step 4: Find the Magnitude of the Relative Velocity
To find the magnitude of the relative velocity vector \( \vec{v_{BA}} = 5 \hat{i} - 5 \hat{j} \), we can use the Pythagorean theorem:
\[ |\vec{v_{BA}}| = \sqrt{(5)^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \text{ km/min} \]

### Step 5: Conclusion
The magnitude of the relative velocity of Car B with respect to Car A is:
\[ \vec{v_{BA}} = 5\sqrt{2} \text{ km/min} \]

### Final Answer
The relative velocity of Car B with respect to Car A is \( 5\sqrt{2} \) km/min.

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