A moving train is brought to rest within ` 20` seconds by applying brakes. Find the initial velocity , if the retardation due to brakes is `2 m//s^(2)`.

Options

A. 40 m/s
B. 50 m/s
C. 60 m/s
D. 70 m/s

Explanation

To solve the problem of finding the initial velocity of a moving train brought to rest within 20 seconds by applying brakes with a retardation of \(2 \, \text{m/s}^2\), we can use the first equation of motion:

\[ V = U + AT \]

where:
- \( V \) is the final velocity,
- \( U \) is the initial velocity,
- \( A \) is the acceleration (retardation in this case, so it will be negative),
- \( T \) is the time.

Given:
- The final velocity \( V = 0 \, \text{m/s} \) (since the train is brought to rest),
- The retardation \( A = -2 \, \text{m/s}^2 \) (negative because it is a deceleration),
- The time \( T = 20 \, \text{s} \).

Now, let's substitute these values into the equation:

\[ 0 = U + (-2) \times 20 \]

Simplify the equation:

\[ 0 = U - 40 \]

To find \( U \), add 40 to both sides of the equation:

\[ U = 40 \, \text{m/s} \]

So, the initial velocity \( U \) of the train is \( 40 \, \text{m/s} \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Final velocity \( V = 0 \, \text{m/s} \)
- Retardation \( A = -2 \, \text{m/s}^2 \)
- Time \( T = 20 \, \text{s} \)

2. **Write the first equation of motion:**
\[ V = U + AT \]

3. **Substitute the given values into the equation:**
\[ 0 = U + (-2) \times 20 \]

4. **Simplify the equation:**
\[ 0 = U - 40 \]

5. **Solve for \( U \):**
\[ U = 40 \, \text{m/s} \]