A moving train is stopped by applying brakes. It stops after travelling `80 m`. If the speed of the train is doubled and retardation remain the same, it will cover a distance-

Options

A. Same as earlier
B. Double the distance covered earlier
C. Four times the distance covered earlier
D. Half the distance covered earlier

Explanation

To solve the problem step by step, we will use the equations of motion and the concept of uniform retardation.

### Step 1: Understand the initial scenario
The train stops after traveling a distance of 80 m when it applies brakes. We denote the initial speed of the train as \( u \) and the retardation (deceleration) as \( a \).

### Step 2: Apply the equation of motion
Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity (0 m/s when the train stops),
- \( u \) is the initial velocity,
- \( a \) is the retardation (negative value),
- \( s \) is the distance traveled (80 m).

Substituting the values, we have:
\[ 0 = u^2 + 2(-a)(80) \]
This simplifies to:
\[ u^2 = 160a \quad \text{(Equation 1)} \]

### Step 3: Consider the new scenario with doubled speed
Now, if the speed of the train is doubled, the new initial speed \( u' \) becomes:
\[ u' = 2u \]
The retardation remains the same, which is \( -a \).

### Step 4: Apply the equation of motion for the new scenario
Using the same equation of motion for the new scenario:
\[ v'^2 = (u')^2 + 2as' \]
where \( v' \) is again 0 (the train stops), and \( s' \) is the new distance traveled. Substituting the values, we have:
\[ 0 = (2u)^2 + 2(-a)s' \]
This simplifies to:
\[ 0 = 4u^2 - 2as' \]
Rearranging gives:
\[ 2as' = 4u^2 \]
Thus:
\[ s' = \frac{4u^2}{2a} = \frac{2u^2}{a} \quad \text{(Equation 2)} \]

### Step 5: Relate the two distances
From Equation 1, we have:
\[ s = \frac{u^2}{2a} = 80 \quad \Rightarrow \quad u^2 = 160a \]
Now, substituting \( u^2 \) from Equation 1 into Equation 2:
\[ s' = \frac{2(160a)}{a} = 320 \text{ m} \]

### Conclusion
The distance covered by the train when its speed is doubled and the retardation remains the same is \( 320 \) m.