A body starts from rest and travels a distance x with uniform acceleration, then it travels a distance 2x with uniform speed, finally it travels a distance 3x with uniform retardation and comes to rest. If the complete motion of the particle is along a straight line, then the ratio of its average velocity to maximum velocity is

Options

A. `(6)/(7)`
B. `(4)/(5)`
C. `(3)/(5)`
D. `(2)/(5)`

Explanation

To solve the problem step by step, we will analyze the motion of the body in three distinct phases: acceleration, constant speed, and retardation.

### Step 1: Analyze the first phase (acceleration)
- The body starts from rest and travels a distance \( x \) with uniform acceleration \( \alpha \).
- Initial velocity \( u = 0 \).
- Final velocity at the end of this phase is \( V' \).
- Using the equation of motion:
\[ V' = u + \alpha t_1 \implies V' = \alpha t_1 \]
- Using the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \implies x = 0 + \frac{1}{2} \alpha t_1^2 \implies x = \frac{1}{2} \alpha t_1^2 \]
- Rearranging gives:
\[ t_1^2 = \frac{2x}{\alpha} \implies t_1 = \sqrt{\frac{2x}{\alpha}} \]

### Step 2: Analyze the second phase (constant speed)
- The body travels a distance \( 2x \) with uniform speed \( V' \).
- The time taken for this phase \( t_2 \) can be calculated as:
\[ s = vt \implies 2x = V' t_2 \implies t_2 = \frac{2x}{V'} \]

### Step 3: Analyze the third phase (retardation)
- The body travels a distance \( 3x \) with uniform retardation \( \beta \) and comes to rest.
- Initial velocity for this phase is \( V' \) and final velocity is \( 0 \).
- Using the equation of motion:
\[ 0 = V' - \beta t_3 \implies \beta = \frac{V'}{t_3} \]
- Using the second equation of motion:
\[ s = vt + \frac{1}{2} a t^2 \implies 3x = V' t_3 - \frac{1}{2} \beta t_3^2 \]
Substituting \( \beta \):
\[ 3x = V' t_3 - \frac{1}{2} \left(\frac{V'}{t_3}\right) t_3^2 \]
Simplifying gives:
\[ 3x = V' t_3 - \frac{1}{2} V' t_3 \implies 3x = \frac{1}{2} V' t_3 \implies t_3 = \frac{6x}{V'} \]

### Step 4: Calculate total time and average velocity
- Total distance traveled:
\[ \text{Total distance} = x + 2x + 3x = 6x \]
- Total time taken:
\[ t_{\text{total}} = t_1 + t_2 + t_3 = \sqrt{\frac{2x}{\alpha}} + \frac{2x}{V'} + \frac{6x}{V'} \]
\[ t_{\text{total}} = \sqrt{\frac{2x}{\alpha}} + \frac{8x}{V'} \]
- Average velocity \( V_{\text{avg}} \):
\[ V_{\text{avg}} = \frac{\text{Total distance}}{\text{Total time}} = \frac{6x}{\sqrt{\frac{2x}{\alpha}} + \frac{8x}{V'}} \]

### Step 5: Calculate the maximum velocity
- The maximum velocity reached during the motion is \( V' \).

### Step 6: Find the ratio of average velocity to maximum velocity
- The ratio of average velocity to maximum velocity is:
\[ \text{Ratio} = \frac{V_{\text{avg}}}{V'} = \frac{6x}{V' \left(\sqrt{\frac{2x}{\alpha}} + \frac{8x}{V'}\right)} \]
- After simplifying, we find that the ratio is \( \frac{3}{5} \) or \( 3:5 \).

### Final Answer
The ratio of the average velocity to the maximum velocity is \( \frac{3}{5} \) or \( 3:5 \).

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