A bus starts from rest and moves whith acceleration `a = 2 m/s^2`. The ratio of the distance covered in `6^(th)` second to that covered in 6 seconds is

Options

A. `11 : 36`
B. `11 : 34`
C. `34 : 11`
D. `36 : 11`

Answer

To solve the problem, we need to find the ratio of the distance covered by the bus in the 6th second to the total distance covered in the first 6 seconds.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity, \( u = 0 \) (the bus starts from rest)
- Acceleration, \( a = 2 \, \text{m/s}^2 \)
- Time interval for which we need to calculate the distances: \( t = 6 \, \text{s} \)

2. **Calculate Distance Covered in the 6th Second:**
- The formula for the distance covered in the \( n^{th} \) second is:
\[ s_n = u + \frac{1}{2} a (2n - 1) \]
- For the 6th second (\( n = 6 \)):
\[ s_6 = 0 + \frac{1}{2} \times 2 \times (2 \times 6 - 1) \]
\[ s_6 = 1 \times (12 - 1) = 1 \times 11 = 11 \, \text{m} \]

3. **Calculate Total Distance Covered in 6 Seconds:**
- The formula for the total distance covered in time \( t \) is:
\[ s = ut + \frac{1}{2} a t^2 \]
- For \( t = 6 \):
\[ s = 0 \times 6 + \frac{1}{2} \times 2 \times 6^2 \]
\[ s = 0 + 1 \times 36 = 36 \, \text{m} \]

4. **Calculate the Ratio:**
- The ratio of the distance covered in the 6th second to the total distance covered in 6 seconds is:
\[ \text{Ratio} = \frac{s_6}{s} = \frac{11}{36} \]

### Final Answer:
The ratio of the distance covered in the 6th second to that covered in 6 seconds is \( \frac{11}{36} \).

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