A train is moving with a velocity of 90 km `h^(-1)`. It is brought to stop by applying the brakes which produce a retardation of 0.5 m `s^(-2) `? Find : (i) the velocity after 10 s, and (ii) the time taken by the train to come to rest.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Given Data:
- Initial velocity of the train, \( u = 90 \, \text{km/h} \)
- Retardation (deceleration), \( a = -0.5 \, \text{m/s}^2 \)
- Time, \( t = 10 \, \text{s} \)

### Step 1: Convert the initial velocity from km/h to m/s
To convert km/h to m/s, we use the conversion factor:
\[ 1 \, \text{km/h} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{1}{3.6} \, \text{m/s} \]
Thus,
\[ u = 90 \, \text{km/h} = 90 \times \frac{1}{3.6} \, \text{m/s} = 25 \, \text{m/s} \]

### Step 2: Calculate the velocity after 10 seconds
We use the equation of motion:
\[ v = u + at \]
Substituting the values:
\[ v = 25 \, \text{m/s} + (-0.5 \, \text{m/s}^2) \times 10 \, \text{s} \]
\[ v = 25 \, \text{m/s} - 5 \, \text{m/s} = 20 \, \text{m/s} \]

### Step 3: Calculate the time taken to come to rest
To find the time taken to come to rest, we set the final velocity \( v = 0 \):
\[ 0 = u + at \]
Substituting the known values:
\[ 0 = 25 \, \text{m/s} + (-0.5 \, \text{m/s}^2) \times t \]
Rearranging gives:
\[ 0.5t = 25 \]
\[ t = \frac{25}{0.5} = 50 \, \text{s} \]

### Final Results:
1. The velocity of the train after 10 seconds is \( 20 \, \text{m/s} \).
2. The time taken by the train to come to rest is \( 50 \, \text{s} \).