A car A moves along north with velocity 30 km/h and another car B moves along east with velocity 40 km/h. The relative velocity of A with respect to B is

Options

A. 50 km/h North - East
B. 50 km/h North-West
C. 50 km/h at angle `tan^(-1)(3//4)` North of West
D. 50 km/h at angle `tan^(-1)(3//4)` West of North

Explanation

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

### Step 1: Identify the velocities of both cars
- Car A moves north with a velocity of 30 km/h.
- Car B moves east with a velocity of 40 km/h.

### Step 2: Represent the velocities as vectors
- The velocity of car A can be represented as:
\[ \vec{V_A} = 30 \, \hat{j} \, \text{km/h} \]
- The velocity of car B can be represented as:
\[ \vec{V_B} = 40 \, \hat{i} \, \text{km/h} \]

### Step 3: Calculate the relative velocity of A with respect to B
The relative velocity of A with respect to B is given by:
\[ \vec{V_{AB}} = \vec{V_A} - \vec{V_B} \]
Substituting the values:
\[ \vec{V_{AB}} = 30 \, \hat{j} - 40 \, \hat{i} \]
This can be rewritten as:
\[ \vec{V_{AB}} = -40 \, \hat{i} + 30 \, \hat{j} \, \text{km/h} \]

### Step 4: Calculate the magnitude of the relative velocity
The magnitude of the relative velocity can be calculated using the Pythagorean theorem:
\[ |\vec{V_{AB}}| = \sqrt{(-40)^2 + (30)^2} \]
Calculating this gives:
\[ |\vec{V_{AB}}| = \sqrt{1600 + 900} = \sqrt{2500} = 50 \, \text{km/h} \]

### Step 5: Determine the direction of the relative velocity
To find the direction, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{30}{40} = \frac{3}{4} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 6: State the final result
The relative velocity of car A with respect to car B is:
\[ 50 \, \text{km/h} \text{ at an angle } \theta = \tan^{-1}\left(\frac{3}{4}\right) \text{ North of West} \]