Man A is sitting in a car moving with a speed of 54 `(km)/(hr)` observes a man B in front of the car crossing perpendicularly a road of width 15 m in three seconds. Then the velocity of man B (in `(m)/(s)`) will be:

Options

A. `5sqrt(10)` towards the car at some angle
B. `5sqrt(10)` away from from the car at some angle
C. 5 perperdicular to the road
D. 15 along the road

Explanation

To solve the problem, we need to find the velocity of man B as observed by man A, who is in a car moving at a speed of 54 km/h. We will follow these steps:

### Step 1: Convert the speed of the car from km/h to m/s
The speed of the car is given as 54 km/h. To convert this to m/s, we use the conversion factor \( \frac{5}{18} \).

\[ \text{Speed of car (m/s)} = 54 \times \frac{5}{18} = 15 \text{ m/s} \]

### Step 2: Determine the speed of man B
Man B crosses a road of width 15 m in 3 seconds. We can calculate his speed using the formula:

\[ \text{Speed of man B} = \frac{\text{Distance}}{\text{Time}} = \frac{15 \text{ m}}{3 \text{ s}} = 5 \text{ m/s} \]

### Step 3: Set up the velocity vectors
Let’s denote:
- The velocity of man A (car) as \( \vec{V_A} = 15 \hat{i} \) m/s (moving in the x-direction).
- The velocity of man B as \( \vec{V_B} = 5 \hat{j} \) m/s (moving in the y-direction).

### Step 4: Calculate the relative velocity of man B with respect to man A
The relative velocity of man B with respect to man A is given by:

\[ \vec{V_{BA}} = \vec{V_B} - \vec{V_A} = 5 \hat{j} - 15 \hat{i} \]

### Step 5: Find the magnitude of the relative velocity
The magnitude of the relative velocity \( \vec{V_{BA}} \) can be calculated using the Pythagorean theorem:

\[ |\vec{V_{BA}}| = \sqrt{(-15)^2 + (5)^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \text{ m/s} \]

### Step 6: Conclusion
Thus, the velocity of man B as observed by man A is \( 5\sqrt{10} \) m/s.