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. Speed of man `B` is `5sqrt(10)m//s`
B. Speed of man `B` is `5ms^(-1)`
C. Actual direction of motion of `B` is at an angle of `tan^(-1)(1/3)` with direction of motion of car
D. Actual direction of motion of `B` is at an angle of `tan^(-1)(3)` with direction opposite to the direction of motion of car.

Answer

To solve the problem, we need to find the velocity of man B as he crosses a road of width 15 meters in 3 seconds while man A is in a car moving at a speed of 54 km/hr. Let's break it down step by step.

### Step 1: Convert the speed of man A from km/hr to m/s
The speed of man A is given as 54 km/hr. To convert this to meters per second, we use the conversion factor:
\[ 1 \text{ km/hr} = \frac{1}{3.6} \text{ m/s} \]
Thus,
\[ 54 \text{ km/hr} = 54 \times \frac{1}{3.6} \text{ m/s} = 15 \text{ m/s} \]

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

### 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 along the x-axis).
- The velocity of man B as \( \vec{V_B} = 5 \hat{j} \) m/s (moving along the y-axis).

### Step 4: Find 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:
\[ \vec{V_{BA}} = \vec{V_B} - \vec{V_A} \]
Substituting the values:
\[ \vec{V_{BA}} = (0 \hat{i} + 5 \hat{j}) - (15 \hat{i} + 0 \hat{j}) \]
\[ \vec{V_{BA}} = -15 \hat{i} + 5 \hat{j} \]

### Step 5: Calculate the magnitude of the velocity of man B
To find the magnitude of the velocity of man B with respect to the ground, we can use the Pythagorean theorem:
\[ |\vec{V_B}| = \sqrt{(15)^2 + (5)^2} \]
Calculating this gives:
\[ |\vec{V_B}| = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \text{ m/s} \]

### Final Answer
Thus, the velocity of man B is \( 5\sqrt{10} \text{ m/s} \).

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