A train is travelling at a speed of `90 km//h`. Brakes are applied so as to produce a uniform acceleration of `- 0.5 m//s^(2)`. Find how far the train will go before it is brought to rest.

Options

A. 484 m
B. 250 m
C. 625 m
D. 400 m

Explanation

To solve the problem, we need to determine the distance the train will travel before coming to a stop after the brakes are applied. We will use the third equation of motion for this purpose.

### Step-by-Step Solution:

1. **Convert the initial speed from km/h to m/s:**
The initial speed \( u \) is given as \( 90 \, \text{km/h} \).
\[ u = 90 \times \frac{5}{18} \, \text{m/s} \]
Simplify the conversion:
\[ u = 90 \times \frac{5}{18} = 25 \, \text{m/s} \]

2. **Identify the final speed and acceleration:**
The final speed \( v \) is \( 0 \, \text{m/s} \) because the train comes to rest.
The acceleration \( a \) is \( -0.5 \, \text{m/s}^2 \) (negative because it is deceleration).

3. **Use the third equation of motion:**
The third equation of motion is:
\[ v^2 - u^2 = 2as \]
Substitute the known values:
\[ 0^2 - 25^2 = 2 \times (-0.5) \times s \]
Simplify the equation:
\[ -625 = -1 \times s \]
\[ s = 625 \, \text{m} \]

### Final Answer:
The train will travel a distance of \( 625 \, \text{m} \) before it is brought to rest.