A bus appears to move northwards at a speed of ` 20sqrt3 " km h"^(-1)` to a man driving his car eastwards with speed ` 20 " km h "^(-1)`. Find the velocity of the bus w.r.t ground.

Answer

To solve the problem of finding the velocity of the bus with respect to the ground, we can follow these steps:

### Step 1: Understand the given data
- The bus appears to move northwards at a speed of \( 20\sqrt{3} \) km/h relative to a man driving his car eastwards at a speed of 20 km/h.
- We need to find the velocity of the bus with respect to the ground.

### Step 2: Define the vectors
- Let the velocity of the bus with respect to the car be denoted as \( \vec{V}_{BC} \).
- The velocity of the bus relative to the car is given as:
\[ \vec{V}_{BC} = 20\sqrt{3} \hat{j} \text{ km/h} \]
(where \( \hat{j} \) represents the north direction).

- Let the velocity of the car be denoted as \( \vec{V}_C \):
\[ \vec{V}_C = 20 \hat{i} \text{ km/h} \]
(where \( \hat{i} \) represents the east direction).

### Step 3: Use the relative velocity formula
According to the relative velocity concept:
\[ \vec{V}_{BC} = \vec{V}_B - \vec{V}_C \]
where \( \vec{V}_B \) is the velocity of the bus with respect to the ground.

Rearranging gives:
\[ \vec{V}_B = \vec{V}_{BC} + \vec{V}_C \]

### Step 4: Substitute the known values
Substituting the values we have:
\[ \vec{V}_B = 20\sqrt{3} \hat{j} + 20 \hat{i} \]

### Step 5: Write the components of the bus's velocity
Thus, the velocity of the bus with respect to the ground can be expressed as:
\[ \vec{V}_B = 20 \hat{i} + 20\sqrt{3} \hat{j} \text{ km/h} \]

### Step 6: Calculate the magnitude of the bus's velocity
The magnitude of \( \vec{V}_B \) can be calculated using the Pythagorean theorem:
\[ |\vec{V}_B| = \sqrt{(20)^2 + (20\sqrt{3})^2} \]
Calculating each term:
\[ |\vec{V}_B| = \sqrt{400 + 1200} = \sqrt{1600} = 40 \text{ km/h} \]

### Step 7: Calculate the direction of the bus's velocity
To find the direction, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{Vertical Component}}{\text{Horizontal Component}} = \frac{20\sqrt{3}}{20} = \sqrt{3} \]
Thus,
\[ \theta = \tan^{-1}(\sqrt{3}) = 60^\circ \]

### Final Result
The velocity of the bus with respect to the ground is:
\[ \vec{V}_B = 40 \text{ km/h at } 60^\circ \text{ from the east direction.} \]

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