A car accelerates uniformly from ` 18 km//h` to `36 km//h` in 5 second. Calculate (i) the acceleration and (ii) the distance covered by the car in that time .

Explanation

To solve the problem step by step, we will calculate the acceleration and the distance covered by the car during the given time.

### Step 1: Convert the speeds from km/h to m/s
The initial speed (u) is given as 18 km/h and the final speed (v) is 36 km/h. We need to convert these speeds into meters per second (m/s) using the conversion factor:
1 km/h = (1/3.6) m/s.

- **Initial speed (u)**:
\[ u = 18 \text{ km/h} = \frac{18}{3.6} \text{ m/s} = 5 \text{ m/s} \]

- **Final speed (v)**:
\[ v = 36 \text{ km/h} = \frac{36}{3.6} \text{ m/s} = 10 \text{ m/s} \]

### Step 2: Calculate the acceleration (a)
We can use the first equation of motion:
\[ v = u + at \]
Rearranging the equation to find acceleration (a):
\[ a = \frac{v - u}{t} \]
Substituting the values:
- \( v = 10 \text{ m/s} \)
- \( u = 5 \text{ m/s} \)
- \( t = 5 \text{ seconds} \)

\[ a = \frac{10 - 5}{5} = \frac{5}{5} = 1 \text{ m/s}^2 \]

### Step 3: Calculate the distance covered (s)
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Substituting the values:
- \( u = 5 \text{ m/s} \)
- \( a = 1 \text{ m/s}^2 \)
- \( t = 5 \text{ seconds} \)

\[ s = (5 \times 5) + \frac{1}{2} \times 1 \times (5^2) \]
Calculating each term:
- First term: \( 5 \times 5 = 25 \)
- Second term: \( \frac{1}{2} \times 1 \times 25 = \frac{25}{2} = 12.5 \)

Adding both terms:
\[ s = 25 + 12.5 = 37.5 \text{ meters} \]

### Final Answers:
(i) The acceleration of the car is \( 1 \text{ m/s}^2 \).

(ii) The distance covered by the car in that time is \( 37.5 \text{ meters} \).

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