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 distance covered in the first 6 seconds. Let's break this down step by step.

### Step 1: Calculate the distance covered in the 6th second

1. **Understanding the motion**: The bus starts from rest and has a constant acceleration \( a = 2 \, \text{m/s}^2 \).
2. **Finding the velocity at the end of the 6th second**:
\[ v = u + at \]
where \( u = 0 \, \text{m/s} \) (initial velocity), \( a = 2 \, \text{m/s}^2 \), and \( t = 6 \, \text{s} \).
\[ v = 0 + (2)(6) = 12 \, \text{m/s} \]
3. **Finding the velocity at the beginning of the 6th second** (which is the end of the 5th second):
\[ v = u + at = 0 + (2)(5) = 10 \, \text{m/s} \]
4. **Calculating the average velocity during the 6th second**:
\[ \text{Average velocity} = \frac{v_{\text{initial}} + v_{\text{final}}}{2} = \frac{10 + 12}{2} = 11 \, \text{m/s} \]
5. **Calculating the distance covered in the 6th second**:
\[ \text{Distance} = \text{Average velocity} \times \text{Time} = 11 \, \text{m/s} \times 1 \, \text{s} = 11 \, \text{m} \]

### Step 2: Calculate the total distance covered in the first 6 seconds

1. **Using the formula for distance under constant acceleration**:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( u = 0 \, \text{m/s} \), \( a = 2 \, \text{m/s}^2 \), and \( t = 6 \, \text{s} \).
\[ s = 0 \times 6 + \frac{1}{2} \times 2 \times (6^2) = 0 + 1 \times 36 = 36 \, \text{m} \]

### Step 3: Calculate the ratio of the distances

1. **Finding the ratio of the distance covered in the 6th second to the distance covered in the first 6 seconds**:
\[ \text{Ratio} = \frac{\text{Distance in 6th second}}{\text{Distance in first 6 seconds}} = \frac{11 \, \text{m}}{36 \, \text{m}} = \frac{11}{36} \]

### Final Answer

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

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