A 100 m long train crosses a man travelling at `5 kmh^(-1)` , in opposite direction, in `7.2 s` then the velocity of train is

Options

A. `40 ms^(-1)`
B. `25 ms^(-1)`
C. `20 ms^(-1)`
D. `45 ms^(-1)`

Explanation

To find the velocity of the train, we can follow these steps:

### Step 1: Understand the problem
We have a train of length 100 meters crossing a man traveling at a speed of 5 km/h in the opposite direction. The time taken to cross the man is 7.2 seconds.

### Step 2: Convert the man's speed to meters per second
The speed of the man is given in km/h. We need to convert this to meters per second (m/s) using the conversion factor:
\[ 1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{1}{3.6} \text{ m/s} \]
So, the speed of the man in m/s is:
\[ V_m = 5 \text{ km/h} \times \frac{1}{3.6} \approx 1.39 \text{ m/s} \]

### Step 3: Set up the equation for relative velocity
When two objects move towards each other, their relative velocity is the sum of their speeds. Let \( V_t \) be the speed of the train in m/s. The relative velocity of the train with respect to the man is:
\[ V_{relative} = V_t + V_m \]

### Step 4: Use the formula for velocity
The formula for velocity is given by:
\[ V_{relative} = \frac{\text{Distance}}{\text{Time}} \]
In this case, the distance is the length of the train (100 m) and the time taken to cross the man is 7.2 s. Therefore:
\[ V_{relative} = \frac{100 \text{ m}}{7.2 \text{ s}} \approx 13.89 \text{ m/s} \]

### Step 5: Set up the equation
Now, we can set up the equation for the relative velocity:
\[ V_t + V_m = 13.89 \text{ m/s} \]
Substituting \( V_m \):
\[ V_t + 1.39 \text{ m/s} = 13.89 \text{ m/s} \]

### Step 6: Solve for the train's velocity
Now, we can solve for \( V_t \):
\[ V_t = 13.89 \text{ m/s} - 1.39 \text{ m/s} \approx 12.50 \text{ m/s} \]

### Step 7: Convert the train's velocity back to km/h
To convert the train's speed back to km/h, we use the conversion factor:
\[ V_t = 12.50 \text{ m/s} \times 3.6 \approx 45 \text{ km/h} \]

### Final Answer
The velocity of the train is approximately **45 km/h**.

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