A particle starts from rest and traverses a distance 2x with uniform acceleration, then moves uniformly over a further distance 4x and finally comes to rest after moving a further distance 6x under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its way is

Options

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

Answer

To solve the problem, we need to analyze the motion of the particle in three segments: acceleration, uniform motion, and deceleration. We will calculate the average speed over the entire journey and the maximum speed reached during the motion.

### Step-by-Step Solution:

1. **Identify the Distances and Segments:**
- The particle moves a distance of \(2x\) with uniform acceleration.
- Then, it moves a distance of \(4x\) with uniform speed.
- Finally, it comes to rest after moving a distance of \(6x\) under uniform retardation.

2. **Calculate the Maximum Speed (\(v_m\)):**
- For the first segment (uniform acceleration), we use the equation of motion:
\[ v_m^2 = u^2 + 2a_1s \]
Here, \(u = 0\), \(s = 2x\), and \(a_1\) is the acceleration.
\[ v_m^2 = 0 + 2a_1(2x) \implies v_m^2 = 4a_1x \]

3. **Calculate the Time for Each Segment:**
- **Time for the first segment (\(t_1\)):**
\[ v_m = u + a_1t_1 \implies v_m = 0 + a_1t_1 \implies t_1 = \frac{v_m}{a_1} \]
- **Time for the second segment (\(t_2\)):**
\[ t_2 = \frac{4x}{v_m} \]
- **Time for the third segment (\(t_3\)):**
- Using the equation of motion:
\[ 0 = v_m + a_3t_3 \implies t_3 = -\frac{v_m}{a_3} \]

4. **Relate the Deceleration (\(a_3\)) to Distance:**
- For the third segment:
\[ 0 = v_m^2 + 2a_3(6x) \implies v_m^2 = -12a_3x \implies a_3 = -\frac{v_m^2}{12x} \]

5. **Substituting \(a_1\) and \(a_3\) into the Time Equations:**
- From \(v_m^2 = 4a_1x\), we can express \(a_1\):
\[ a_1 = \frac{v_m^2}{4x} \]
- Substitute \(a_1\) into \(t_1\):
\[ t_1 = \frac{v_m}{\frac{v_m^2}{4x}} = \frac{4x}{v_m} \]
- Substitute \(a_3\) into \(t_3\):
\[ t_3 = -\frac{v_m}{-\frac{v_m^2}{12x}} = \frac{12x}{v_m} \]

6. **Calculate Total Time (\(T\)):**
\[ T = t_1 + t_2 + t_3 = \frac{4x}{v_m} + \frac{4x}{v_m} + \frac{12x}{v_m} = \frac{20x}{v_m} \]

7. **Calculate Total Distance:**
\[ \text{Total Distance} = 2x + 4x + 6x = 12x \]

8. **Calculate Average Speed (\(v_{avg}\)):**
\[ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{12x}{\frac{20x}{v_m}} = \frac{12v_m}{20} = \frac{3v_m}{5} \]

9. **Calculate the Ratio of Average Speed to Maximum Speed:**
\[ \text{Ratio} = \frac{v_{avg}}{v_m} = \frac{\frac{3v_m}{5}}{v_m} = \frac{3}{5} \]

### Final Answer:
The ratio of average speed over the journey to the maximum speed on its way is \(\frac{3}{5}\).