When brakes are applied to a bus, the retardation produced is 25 cm `s^(-2)` and the bus takes 20 s to stop. Calculate : (i) the initial velocity of bus, and (ii) the distance travelled by bus during this time.

Explanation

To solve the problem, we will break it down into two parts: (i) calculating the initial velocity of the bus and (ii) calculating the distance traveled by the bus during the time it takes to stop.

### Given Data:
- Retardation (a) = 25 cm/s² (which is negative acceleration)
- Time (t) = 20 s
- Final velocity (v) = 0 cm/s (since the bus comes to a stop)

### Step 1: Calculate the Initial Velocity (u)

Using the first equation of motion:
\[ v = u + at \]

Where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration (retardation in this case, so it will be negative)
- \( t \) = time

Substituting the known values:
\[ 0 = u - 25 \times 20 \]

Rearranging the equation to solve for \( u \):
\[ u = 25 \times 20 \]
\[ u = 500 \text{ cm/s} \]

### Step 2: Calculate the Distance Traveled (s)

Using the second equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]

Substituting the values we have:
\[ s = (500 \text{ cm/s}) \times (20 \text{ s}) + \frac{1}{2} \times (-25 \text{ cm/s}^2) \times (20 \text{ s})^2 \]

Calculating each term:
1. First term:
\[ 500 \times 20 = 10000 \text{ cm} \]
2. Second term:
\[ \frac{1}{2} \times (-25) \times 400 = -5000 \text{ cm} \]

Now, substituting back into the equation for \( s \):
\[ s = 10000 \text{ cm} - 5000 \text{ cm} \]
\[ s = 5000 \text{ cm} \]

### Convert Distance to Meters
Since \( 1 \text{ cm} = 0.01 \text{ m} \):
\[ s = 5000 \text{ cm} \times 0.01 \text{ m/cm} = 50 \text{ m} \]

### Final Answers:
(i) The initial velocity of the bus is **500 cm/s**.
(ii) The distance traveled by the bus during this time is **50 m**.

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