A body starts from rest and travels a distance `S` with unitorm acceleration, then moves uniformly a distance `2S` uniformly , and finally cones to rest after moving further `5S` under uniform retardation. The ratio of the average velocity to maximum velocity is.

Options

A. a.` 2//5`
B. b. ` 3//5`
C. c. ` 4//7`
D. d. ` 5//7`

Explanation

To solve the problem step by step, we will analyze the motion of the body in three distinct phases: acceleration, uniform motion, and deceleration. We will then find the average velocity and the maximum velocity to determine their ratio.

### Step 1: Analyze the phases of motion

1. **Phase 1 (Uniform Acceleration)**:
- The body starts from rest and travels a distance \( S \) with uniform acceleration.
- Initial velocity \( u = 0 \).
- Distance covered \( S \).
- Let the acceleration be \( a \).
- Using the equation of motion:
\[ S = ut + \frac{1}{2} a t_1^2 \implies S = 0 + \frac{1}{2} a t_1^2 \implies t_1^2 = \frac{2S}{a} \implies t_1 = \sqrt{\frac{2S}{a}} \]

2. **Phase 2 (Uniform Motion)**:
- The body moves uniformly for a distance \( 2S \) after accelerating.
- The maximum velocity \( v_{max} \) at the end of phase 1 is given by:
\[ v_{max} = u + at_1 = 0 + a t_1 = a \sqrt{\frac{2S}{a}} = \sqrt{2aS} \]
- Time taken to cover \( 2S \) at constant velocity \( v_{max} \):
\[ t_2 = \frac{2S}{v_{max}} = \frac{2S}{\sqrt{2aS}} = \frac{2\sqrt{S}}{\sqrt{2a}} = \frac{2\sqrt{S}}{\sqrt{2a}} \]

3. **Phase 3 (Uniform Deceleration)**:
- The body comes to rest after moving a distance \( 5S \) under uniform retardation.
- Using the equation of motion for deceleration:
\[ 5S = v_{max} t_3 - \frac{1}{2} a' t_3^2 \]
- Here, \( a' \) is the retardation. Since it comes to rest, we can express \( t_3 \) in terms of \( v_{max} \) and \( a' \):
\[ 5S = v_{max} t_3 - \frac{1}{2} a' t_3^2 \]
- We can also express \( a' \) in terms of \( v_{max} \) and \( t_3 \):
\[ v_{max} = a' t_3 \implies a' = \frac{v_{max}}{t_3} \]

### Step 2: Calculate Total Distance and Total Time

- Total distance traveled:
\[ S + 2S + 5S = 8S \]

- Total time taken:
\[ t_{total} = t_1 + t_2 + t_3 \]
- Substitute \( t_1 \), \( t_2 \), and \( t_3 \) into the equation.

### Step 3: Calculate Average Velocity

- Average velocity \( v_{avg} \) is given by:
\[ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{8S}{t_{total}} \]

### Step 4: Calculate the Ratio of Average Velocity to Maximum Velocity

- The maximum velocity \( v_{max} \) was calculated as \( \sqrt{2aS} \).
- The ratio of average velocity to maximum velocity is:
\[ \frac{v_{avg}}{v_{max}} = \frac{\frac{8S}{t_{total}}}{v_{max}} \]

### Final Calculation

1. Substitute the values of \( t_{total} \) and \( v_{max} \) into the ratio.
2. Simplify the expression to find the ratio.

### Conclusion

The final ratio of the average velocity to the maximum velocity is:
\[ \frac{v_{avg}}{v_{max}} = \frac{4}{7} \]