A train 110 m long passes a man running at a speed of 6km/hr in the direction opposite to the train in 6 seconds. What is the speed of the train?

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Problem
We have a train that is 110 meters long, and it passes a man running in the opposite direction at a speed of 6 km/hr in 6 seconds. We need to find the speed of the train.

### Step 2: Convert the Man's Speed to Meters per Second
The speed of the man is given in kilometers per hour (km/hr). We need to convert this to meters per second (m/s) using the conversion factor:
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
So, the speed of the man in m/s is:
\[ \text{Speed of man} = 6 \text{ km/hr} \times \frac{5}{18} = \frac{30}{18} = \frac{5}{3} \text{ m/s} \]

### Step 3: Calculate the Relative Speed
Since the train and the man are moving in opposite directions, their speeds will add up. Let the speed of the train be \( x \) km/hr. In meters per second, the speed of the train is:
\[ \text{Speed of train} = x \text{ km/hr} \times \frac{5}{18} \text{ m/s} \]
The relative speed (combined speed) when they are moving in opposite directions is:
\[ \text{Relative speed} = \text{Speed of train} + \text{Speed of man} \]
In m/s, this becomes:
\[ \text{Relative speed} = \left( x \times \frac{5}{18} + \frac{5}{3} \right) \]

### Step 4: Set Up the Equation Using Distance and Time
The distance covered by the train while passing the man is equal to the length of the train, which is 110 meters. The time taken to pass the man is 6 seconds. Using the formula:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]
we can write:
\[ 110 = \left( x \times \frac{5}{18} + \frac{5}{3} \right) \times 6 \]

### Step 5: Simplify the Equation
First, simplify the right side:
\[ 110 = 6 \left( x \times \frac{5}{18} + \frac{5}{3} \right) \]
Dividing both sides by 6:
\[ \frac{110}{6} = x \times \frac{5}{18} + \frac{5}{3} \]
Calculating \( \frac{110}{6} \):
\[ \frac{110}{6} = \frac{55}{3} \]
So we have:
\[ \frac{55}{3} = x \times \frac{5}{18} + \frac{5}{3} \]
Now, subtract \( \frac{5}{3} \) from both sides:
\[ \frac{55}{3} - \frac{5}{3} = x \times \frac{5}{18} \]
This simplifies to:
\[ \frac{50}{3} = x \times \frac{5}{18} \]

### Step 6: Solve for \( x \)
Multiply both sides by \( \frac{18}{5} \):
\[ x = \frac{50}{3} \times \frac{18}{5} \]
Calculating this gives:
\[ x = \frac{50 \times 18}{3 \times 5} = \frac{900}{15} = 60 \text{ km/hr} \]

### Final Answer
The speed of the train is **60 km/hr**.

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