A bus is moving with a velocity of `40 kmh^(-1)` due south, and a car is moving due west with a velocity of `50kmh^(-1)`. What is the velocity of car w.r.t. a passenger in the bus ?

Answer

To find the velocity of the car with respect to a passenger in the bus, we can follow these steps:

### Step 1: Identify the velocities
- The bus is moving due south with a velocity of \( V_b = 40 \, \text{km/h} \).
- The car is moving due west with a velocity of \( V_c = 50 \, \text{km/h} \).

### Step 2: Set up a coordinate system
- Let's define a coordinate system:
- The positive y-axis points north.
- The negative y-axis points south.
- The positive x-axis points east.
- The negative x-axis points west.

### Step 3: Represent the velocities as vectors
- The velocity of the bus can be represented as:
\[ \vec{V_b} = 0 \hat{i} - 40 \hat{j} \, \text{(since it is moving south)} \]
- The velocity of the car can be represented as:
\[ \vec{V_c} = -50 \hat{i} + 0 \hat{j} \, \text{(since it is moving west)} \]

### Step 4: Calculate the relative velocity of the car with respect to the bus
- The velocity of the car with respect to the bus is given by:
\[ \vec{V_{cb}} = \vec{V_c} - \vec{V_b} \]
- Substituting the values:
\[ \vec{V_{cb}} = (-50 \hat{i} + 0 \hat{j}) - (0 \hat{i} - 40 \hat{j}) = -50 \hat{i} + 40 \hat{j} \]

### Step 5: Calculate the magnitude of the relative velocity
- The magnitude of the relative velocity can be calculated using the Pythagorean theorem:
\[ |\vec{V_{cb}}| = \sqrt{(-50)^2 + (40)^2} \]
- Calculating the squares:
\[ |\vec{V_{cb}}| = \sqrt{2500 + 1600} = \sqrt{4100} \]
- Simplifying further:
\[ |\vec{V_{cb}}| \approx 64.03 \, \text{km/h} \]

### Step 6: Determine the direction of the relative velocity
- The direction can be found using the tangent function:
\[ \tan(\theta) = \frac{40}{50} \]
- Therefore:
\[ \theta = \tan^{-1}\left(\frac{40}{50}\right) \approx 38.66^\circ \, \text{(north of west)} \]

### Final Answer
The velocity of the car with respect to a passenger in the bus is approximately \( 64.03 \, \text{km/h} \) at an angle of \( 38.66^\circ \) north of west.

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