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 solve the problem, we need to find the velocity of the train given the information about the man and the time taken for the train to cross him. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Length of the train (L) = 100 m
- Speed of the man (Vm) = 5 km/h
- Time taken to cross the man (t) = 7.2 s

### Step 2: Convert the speed of the man from km/h to m/s
To convert km/h to m/s, we use the conversion factor \( \frac{5}{18} \):
\[ Vm = 5 \text{ km/h} \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]

### Step 3: Set up the equation for relative velocity
Since the train and the man are moving in opposite directions, the relative velocity (Vr) of the train with respect to the man is given by:
\[ Vr = Vt + Vm \]
Where \( Vt \) is the velocity of the train.

### Step 4: Use the formula for distance
The distance traveled by the train while crossing the man is equal to the length of the train:
\[ \text{Distance} = \text{Relative Velocity} \times \text{Time} \]
Thus,
\[ 100 = (Vt + \frac{25}{18}) \times 7.2 \]

### Step 5: Rearrange the equation to solve for Vt
First, divide both sides by 7.2:
\[ \frac{100}{7.2} = Vt + \frac{25}{18} \]
Calculating \( \frac{100}{7.2} \):
\[ \frac{100}{7.2} \approx 13.89 \]
Now, we have:
\[ 13.89 = Vt + \frac{25}{18} \]

### Step 6: Convert \( \frac{25}{18} \) to decimal
Calculating \( \frac{25}{18} \):
\[ \frac{25}{18} \approx 1.39 \]

### Step 7: Substitute and solve for Vt
Now substitute back into the equation:
\[ 13.89 = Vt + 1.39 \]
Subtract \( 1.39 \) from both sides:
\[ Vt = 13.89 - 1.39 = 12.5 \text{ m/s} \]

### Step 8: Convert the velocity of the train back to km/h
To convert m/s back to km/h, multiply by \( 18/5 \):
\[ Vt = 12.5 \text{ m/s} \times \frac{18}{5} = 45 \text{ km/h} \]

### Final Answer
The velocity of the train is \( 45 \text{ km/h} \).

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