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 determine the velocity of man B as observed by man A in the car. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert the speed of the car from km/hr to m/s
The speed of the car (man A) is given as 54 km/hr. To convert this to meters per second (m/s), we use the conversion factor:
\[ \text{Speed in m/s} = \text{Speed in km/hr} \times \frac{5}{18} \]
Calculating this gives:
\[ 54 \times \frac{5}{18} = 15 \text{ m/s} \]

### Step 2: Calculate the speed of man B
Man B crosses a road of width 15 m in 3 seconds. The speed of man B can be calculated using the formula:
\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \]
Substituting the values:
\[ \text{Speed of B} = \frac{15 \text{ m}}{3 \text{ s}} = 5 \text{ m/s} \]

### Step 3: Establish the relative velocities
Let’s denote:
- \( v_A = 15 \text{ m/s} \) (velocity of man A)
- \( v_B = 5 \text{ m/s} \) (velocity of man B)

Since man B is moving perpendicular to the direction of man A, we can represent the velocities as vectors. The velocity of man B is in the positive y-direction, while the velocity of man A is in the positive x-direction.

### Step 4: Set up the equation for relative velocity
The relative velocity of man B with respect to man A is given by:
\[ v_{B/A} = v_B - v_A \]
In vector form, this can be expressed as:
\[ v_{B/A} = (0 \hat{i} + 5 \hat{j}) - (15 \hat{i} + 0 \hat{j}) = -15 \hat{i} + 5 \hat{j} \]

### Step 5: Calculate the magnitude of the relative velocity
The magnitude of the relative velocity vector can be calculated using the Pythagorean theorem:
\[ |v_{B/A}| = \sqrt{(-15)^2 + (5)^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \text{ m/s} \]

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