A body starts from rest and travels a distance `S` with uniform 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. ` 2//5`
B. ` 3//5`
C. ` 4//7`
D. ` 5//7`

Explanation

To solve the problem step by step, we will analyze the motion of the body in three parts: acceleration, uniform motion, and deceleration.

### Step 1: Analyze the first part of the motion (Uniform Acceleration)
- The body starts from rest and travels a distance \( S \) with uniform acceleration.
- Initial velocity \( u = 0 \).
- Let the acceleration be \( a \).
- Using the equation of motion:
\[ v^2 = u^2 + 2aS \]
Substituting \( u = 0 \):
\[ v^2 = 2aS \quad \text{(1)} \]
Here, \( v \) is the final velocity after traveling distance \( S \).

### Step 2: Calculate the time taken in the first part
- Using the formula for time under uniform acceleration:
\[ S = ut + \frac{1}{2}at^2 \]
Substituting \( u = 0 \):
\[ S = \frac{1}{2}at_1^2 \]
Rearranging gives:
\[ t_1 = \sqrt{\frac{2S}{a}} \quad \text{(2)} \]

### Step 3: Analyze the second part of the motion (Uniform Velocity)
- The body moves a distance \( 2S \) with constant velocity \( v \).
- The time taken for this part is:
\[ t_2 = \frac{2S}{v} \quad \text{(3)} \]

### Step 4: Analyze the third part of the motion (Uniform Retardation)
- The body comes to rest after moving a distance \( 5S \) with uniform retardation. Let the retardation be \( a_1 \).
- Using the equation of motion:
\[ 0 = v^2 - 2a_1(5S) \]
Rearranging gives:
\[ a_1 = \frac{v^2}{10S} \quad \text{(4)} \]

### Step 5: Calculate the time taken in the third part
- Using the formula for time under uniform retardation:
\[ 0 = v - a_1 t_3 \]
Rearranging gives:
\[ t_3 = \frac{v}{a_1} \]
Substituting \( a_1 \) from equation (4):
\[ t_3 = \frac{v}{\frac{v^2}{10S}} = \frac{10S}{v} \quad \text{(5)} \]

### Step 6: Calculate total displacement and total time
- Total displacement \( = S + 2S + 5S = 8S \).
- Total time \( = t_1 + t_2 + t_3 \):
\[ t_{total} = \sqrt{\frac{2S}{a}} + \frac{2S}{v} + \frac{10S}{v} \]
Simplifying gives:
\[ t_{total} = \sqrt{\frac{2S}{a}} + \frac{12S}{v} \quad \text{(6)} \]

### Step 7: Calculate average velocity
- Average velocity \( V_{avg} \) is given by:
\[ V_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{8S}{t_{total}} \]
Substituting from (6):
\[ V_{avg} = \frac{8S}{\sqrt{\frac{2S}{a}} + \frac{12S}{v}} \quad \text{(7)} \]

### Step 8: Calculate maximum velocity
- The maximum velocity \( V_{max} = v \) from the first part.

### Step 9: Calculate the ratio of average velocity to maximum velocity
- The ratio \( \frac{V_{avg}}{V_{max}} \):
\[ \frac{V_{avg}}{V_{max}} = \frac{\frac{8S}{\sqrt{\frac{2S}{a}} + \frac{12S}{v}}}{v} \]
Simplifying gives:
\[ = \frac{8S}{v(\sqrt{\frac{2S}{a}} + \frac{12S}{v})} \]

### Final Calculation
Using the relationships derived, we find that the ratio simplifies to:
\[ \frac{8}{14} = \frac{4}{7} \]

### Conclusion
The ratio of the average velocity to the maximum velocity is \( \frac{4}{7} \).