A car A is moving with speed `"40 km h"^(-1)` along a straight line `30^(@)` north of east and another car B is moving with same speed along a straight line `30^(@)` south of east. The relative velocity of car A as observed from the car B is

Options

A. `"40 km h"^(-1)" north - east"`
B. `"40 km h"^(-1)" south"`
C. `"40 km h"^(-1)" north"`
D. `"40 km h"^(-1)" south - east"`

Answer

To solve the problem of finding the relative velocity of car A as observed from car B, we can follow these steps:

### Step 1: Define the velocities of the cars in vector form.
Car A is moving at a speed of 40 km/h at an angle of 30 degrees north of east. We can break this velocity into its components:

- The eastward (x-direction) component:
\[ V_{A_x} = 40 \cos(30^\circ) \]
- The northward (y-direction) component:
\[ V_{A_y} = 40 \sin(30^\circ) \]

Thus, the velocity vector of car A can be expressed as:
\[ \vec{V_A} = V_{A_x} \hat{i} + V_{A_y} \hat{j} = 40 \cos(30^\circ) \hat{i} + 40 \sin(30^\circ) \hat{j} \]

### Step 2: Calculate the components of car A's velocity.
Using the trigonometric values:
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
- \(\sin(30^\circ) = \frac{1}{2}\)

We can substitute these values:
\[ V_{A_x} = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} \text{ km/h} \]
\[ V_{A_y} = 40 \cdot \frac{1}{2} = 20 \text{ km/h} \]

So,
\[ \vec{V_A} = 20\sqrt{3} \hat{i} + 20 \hat{j} \]

### Step 3: Define the velocity of car B in vector form.
Car B is also moving at a speed of 40 km/h but at an angle of 30 degrees south of east. The components of car B's velocity are:

- The eastward (x-direction) component:
\[ V_{B_x} = 40 \cos(30^\circ) = 20\sqrt{3} \text{ km/h} \]
- The southward (y-direction) component (note that this is negative):
\[ V_{B_y} = -40 \sin(30^\circ) = -20 \text{ km/h} \]

Thus, the velocity vector of car B can be expressed as:
\[ \vec{V_B} = V_{B_x} \hat{i} + V_{B_y} \hat{j} = 20\sqrt{3} \hat{i} - 20 \hat{j} \]

### Step 4: Calculate the relative velocity of car A with respect to car B.
The relative velocity of A with respect to B is given by:
\[ \vec{V_{AB}} = \vec{V_A} - \vec{V_B} \]
Substituting the vectors we found:
\[ \vec{V_{AB}} = (20\sqrt{3} \hat{i} + 20 \hat{j}) - (20\sqrt{3} \hat{i} - 20 \hat{j}) \]

### Step 5: Simplify the expression.
When we perform the subtraction:
\[ \vec{V_{AB}} = 20\sqrt{3} \hat{i} + 20 \hat{j} - 20\sqrt{3} \hat{i} + 20 \hat{j} \]
The \( \hat{i} \) components cancel out:
\[ \vec{V_{AB}} = 0 \hat{i} + (20 + 20) \hat{j} = 40 \hat{j} \]

### Step 6: Interpret the result.
The relative velocity of car A as observed from car B is:
\[ \vec{V_{AB}} = 40 \hat{j} \text{ km/h} \]
This indicates that car A is moving north at a speed of 40 km/h relative to car B.

### Final Answer:
The relative velocity of car A as observed from car B is \( 40 \text{ km/h north} \).

---