A train starting from rest accelerates uniformly for 100 s, then comes to a stop with a uniform retardation in the next 200 s. During the motion, it covers a distance of 3 km. Choose the wrong option

Options

A. Its acceleration is `0.2 m//s^(2)`
B. Its retardation is `0.1 m//s^(2)`
C. The maximum velocity is `20 m//s`
D. The maximum velocity is `10 m//s`

Explanation

To solve the problem step by step, we will analyze the motion of the train, which involves uniform acceleration followed by uniform retardation.

### Step 1: Understanding the Motion
The train starts from rest, accelerates uniformly for 100 seconds, and then comes to a stop with uniform retardation over the next 200 seconds. The total distance covered during this motion is 3 km (or 3000 meters).

### Step 2: Finding Maximum Velocity
1. **Acceleration Phase:**
- Initial velocity (u) = 0 (since it starts from rest)
- Time of acceleration (t1) = 100 seconds
- Let the acceleration be 'a'.
- Final velocity after acceleration (V) = u + a * t1 = 0 + a * 100 = 100a.

2. **Distance covered during acceleration (S1):**
- Using the formula for distance under uniform acceleration:
\[ S_1 = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} a (100)^2 = 5000a. \]

### Step 3: Retardation Phase
1. **Retardation Phase:**
- Initial velocity (u) = V = 100a (from the previous phase).
- Final velocity (V) = 0 (since it comes to a stop).
- Time of retardation (t2) = 200 seconds.
- Let the retardation be 'b'.
- Using the equation of motion:
\[ V = u - b * t2 \Rightarrow 0 = 100a - b * 200 \Rightarrow b = \frac{100a}{200} = 0.5a. \]

2. **Distance covered during retardation (S2):**
- Using the formula for distance under uniform retardation:
\[ S_2 = ut - \frac{1}{2} b t^2 = 100a * 200 - \frac{1}{2} (0.5a) (200)^2. \]
- Simplifying this:
\[ S_2 = 20000a - \frac{1}{2} (0.5a) (40000) = 20000a - 10000a = 10000a. \]

### Step 4: Total Distance
The total distance covered (S) is the sum of distances during acceleration and retardation:
\[ S = S_1 + S_2 = 5000a + 10000a = 15000a. \]
Given that the total distance is 3000 meters:
\[ 15000a = 3000 \Rightarrow a = \frac{3000}{15000} = 0.2 \text{ m/s}^2. \]

### Step 5: Finding Maximum Velocity
Now substituting 'a' back to find the maximum velocity:
\[ V = 100a = 100 * 0.2 = 20 \text{ m/s}. \]

### Step 6: Finding Retardation
Using the value of 'a' to find the retardation:
\[ b = 0.5a = 0.5 * 0.2 = 0.1 \text{ m/s}^2. \]

### Conclusion
From the calculations:
- Maximum velocity (V) = 20 m/s
- Acceleration (a) = 0.2 m/s²
- Retardation (b) = 0.1 m/s²

### Identifying the Wrong Option
The wrong option would be any statement that contradicts these calculated values, such as stating that the maximum velocity is 10 m/s or that the acceleration is different from 0.2 m/s².