The driver of a train moving at `72 km h^(-1)` sights another train moving at `4 ms^(-1)` on the same track and in the same direction. He instantly applies brakes to produces a retardation of `1 ms^(-2)`. The minimum distance between the trains so that no collision occurs is

Options

A. 32 m
B. 64 m
C. 128 m
D. 256 m

Answer

To solve the problem, we need to find the minimum distance between the two trains so that no collision occurs. Here’s a step-by-step breakdown of the solution:

### Step 1: Convert the speed of the first train from km/h to m/s
The speed of the first train is given as \(72 \, \text{km/h}\). We can convert this to meters per second using the conversion factor \(1 \, \text{km/h} = \frac{5}{18} \, \text{m/s}\).

\[ \text{Speed of first train} = 72 \times \frac{5}{18} = 20 \, \text{m/s} \]

### Step 2: Identify the speeds of both trains
- Speed of the first train (Train A): \(20 \, \text{m/s}\)
- Speed of the second train (Train B): \(4 \, \text{m/s}\)

### Step 3: Calculate the relative speed
Since both trains are moving in the same direction, the relative speed of Train A with respect to Train B is:

\[ \text{Relative speed} = \text{Speed of Train A} - \text{Speed of Train B} = 20 \, \text{m/s} - 4 \, \text{m/s} = 16 \, \text{m/s} \]

### Step 4: Use the equation of motion to find the stopping distance
The first train applies brakes with a retardation of \(1 \, \text{m/s}^2\). We can use the equation of motion:

\[ v^2 = u^2 + 2as \]

Where:
- \(v\) = final velocity (0 m/s, since we want to stop)
- \(u\) = initial velocity (relative speed, \(16 \, \text{m/s}\))
- \(a\) = acceleration (retardation, \(-1 \, \text{m/s}^2\))
- \(s\) = distance (what we want to find)

Substituting the values:

\[ 0 = (16)^2 + 2(-1)s \]
\[ 0 = 256 - 2s \]
\[ 2s = 256 \]
\[ s = \frac{256}{2} = 128 \, \text{m} \]

### Step 5: Conclusion
The minimum distance between the two trains to avoid a collision is \(128 \, \text{m}\).

---