Particle beginning from rest, travels a distance `S` with uniform acceleration and immediately after travels a distane of `3 S` with uniform velocity followed by a distance `5 S` with uniform deceleration, and comes to rest. Then the ratio of average speed to the maximum speed of the particle is.

Options

A. `1//5`
B. `2//5`
C. `(3)/(5)`
D. `(4)/(5)`

Explanation

To solve the problem step by step, we will analyze the motion of the particle in three distinct phases: acceleration, uniform velocity, and deceleration.

### Step 1: Analyze the Motion Phases
1. **Phase 1 (Acceleration)**: The particle starts from rest and travels a distance \( S \) with uniform acceleration.
- Initial velocity \( u = 0 \)
- Distance \( s_1 = S \)
- Let the acceleration be \( a \) and the time taken be \( t_1 \).
- Using the equation of motion:
\[ S = ut_1 + \frac{1}{2} a t_1^2 \implies S = 0 + \frac{1}{2} a t_1^2 \implies a = \frac{2S}{t_1^2} \]
- The final velocity at the end of this phase (maximum velocity) is given by:
\[ v_{\text{max}} = u + at_1 = 0 + a t_1 = a t_1 = \frac{2S}{t_1^2} t_1 = \frac{2S}{t_1} \]

### Step 2: Uniform Velocity Phase
2. **Phase 2 (Uniform Velocity)**: The particle travels a distance of \( 3S \) with uniform velocity \( v_{\text{max}} \).
- Distance \( s_2 = 3S \)
- Time taken \( t_2 \) can be calculated as:
\[ t_2 = \frac{s_2}{v_{\text{max}}} = \frac{3S}{\frac{2S}{t_1}} = \frac{3S \cdot t_1}{2S} = \frac{3t_1}{2} \]

### Step 3: Deceleration Phase
3. **Phase 3 (Deceleration)**: The particle travels a distance of \( 5S \) and comes to rest.
- Distance \( s_3 = 5S \)
- Let the time taken for this phase be \( t_3 \).
- Using the equation of motion:
\[ s_3 = v_{\text{max}} t_3 - \frac{1}{2} a' t_3^2 \]
- Here, \( a' \) is the deceleration. Since the particle comes to rest, we can express the deceleration in terms of \( v_{\text{max}} \) and \( t_3 \):
\[ 5S = v_{\text{max}} t_3 - \frac{1}{2} a' t_3^2 \]
- The deceleration can also be expressed as \( a' = \frac{v_{\text{max}}}{t_3} \).

### Step 4: Total Distance and Time
4. **Total Distance**: The total distance traveled by the particle is:
\[ S + 3S + 5S = 9S \]

5. **Total Time**: The total time taken is:
\[ t_{\text{total}} = t_1 + t_2 + t_3 = t_1 + \frac{3t_1}{2} + t_3 = \frac{5t_1}{2} + t_3 \]

### Step 5: Average Speed Calculation
6. **Average Speed**: The average speed \( v_{\text{avg}} \) is given by:
\[ v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{9S}{t_{\text{total}}} \]

### Step 6: Maximum Speed Calculation
7. **Maximum Speed**: The maximum speed \( v_{\text{max}} \) is given by:
\[ v_{\text{max}} = \frac{2S}{t_1} \]

### Step 7: Ratio of Average Speed to Maximum Speed
8. **Ratio**:
\[ \text{Ratio} = \frac{v_{\text{avg}}}{v_{\text{max}}} = \frac{\frac{9S}{t_{\text{total}}}}{\frac{2S}{t_1}} = \frac{9S}{t_{\text{total}}} \cdot \frac{t_1}{2S} = \frac{9t_1}{2t_{\text{total}}} \]

### Step 8: Substitute \( t_{\text{total}} \)
9. Substitute \( t_{\text{total}} = \frac{5t_1}{2} + t_3 \) into the ratio and simplify to find the final answer.

### Final Result
After simplifying, we find that the ratio of average speed to maximum speed is \( \frac{3}{5} \).