A particle starts from rest and traverses a distance l with uniform acceleration, then moves uniformly over a further distance 2l and finally comes to rest after moving a further distance 3l under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its ways 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 segments: acceleration, uniform motion, and deceleration.

### Step 1: Understanding the Motion
The particle undergoes three phases:
1. **Acceleration**: From rest to maximum speed while covering a distance \( l \).
2. **Uniform Motion**: Moving at maximum speed over a distance \( 2l \).
3. **Deceleration**: Coming to rest while covering a distance \( 3l \).

### Step 2: Calculate Average Speed
The average speed \( V_{avg} \) is given by the formula:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} \]
The total distance covered by the particle is:
\[ \text{Total Distance} = l + 2l + 3l = 6l \]

### Step 3: Calculate Time for Each Segment
1. **For the first segment (Acceleration)**:
- Using the equation of motion:
\[ l = \frac{1}{2} a T_1^2 \quad \text{(where \( a \) is acceleration and \( T_1 \) is time taken)} \]
- The maximum speed \( V_{max} \) at the end of this segment is:
\[ V_{max} = a T_1 \]
- Rearranging gives:
\[ T_1 = \frac{2l}{V_{max}} \]

2. **For the second segment (Uniform Motion)**:
- The time taken \( T_2 \) is:
\[ T_2 = \frac{2l}{V_{max}} \]

3. **For the third segment (Deceleration)**:
- Using the equation of motion:
\[ 3l = V_{max} T_3 - \frac{1}{2} b T_3^2 \quad \text{(where \( b \) is retardation)} \]
- Since the particle comes to rest, we can express \( T_3 \) in terms of \( V_{max} \):
\[ T_3 = \frac{6l}{V_{max}} \]

### Step 4: Total Time Calculation
Now, we can sum the times for all three segments:
\[ T_{total} = T_1 + T_2 + T_3 = \frac{2l}{V_{max}} + \frac{2l}{V_{max}} + \frac{6l}{V_{max}} = \frac{10l}{V_{max}} \]

### Step 5: Calculate Average Speed
Now we can substitute the total distance and total time into the average speed formula:
\[ V_{avg} = \frac{6l}{\frac{10l}{V_{max}}} = \frac{6l \cdot V_{max}}{10l} = \frac{6}{10} V_{max} = \frac{3}{5} V_{max} \]

### Step 6: Ratio of Average Speed to Maximum Speed
The ratio of average speed to maximum speed is:
\[ \frac{V_{avg}}{V_{max}} = \frac{\frac{3}{5} V_{max}}{V_{max}} = \frac{3}{5} \]

### Final Answer
Thus, the ratio of average speed over the journey to the maximum speed on its way is:
\[ \frac{3}{5} \]