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 distinct phases: acceleration, uniform motion, and deceleration.

### Step 1: Analyze the first phase (acceleration)
The particle starts from rest and travels a distance \( l \) with uniform acceleration.

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where \( s = l \), \( u = 0 \) (starts from rest), we have:
\[ l = \frac{1}{2} a_1 t_1^2 \]
From this, we can express \( t_1 \):
\[ t_1 = \sqrt{\frac{2l}{a_1}} \]

### Step 2: Analyze the second phase (uniform motion)
The particle then moves uniformly over a distance \( 2l \) with maximum speed \( V_m \).

The time taken \( t_2 \) for this distance is:
\[ t_2 = \frac{2l}{V_m} \]

### Step 3: Analyze the third phase (deceleration)
Finally, the particle comes to rest after moving a distance \( 3l \) under uniform retardation.

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where \( v = 0 \) (final velocity), \( u = V_m \), and \( s = 3l \):
\[ 0 = V_m^2 - 2a_3(3l) \]
From this, we can express \( a_3 \):
\[ a_3 = \frac{V_m^2}{6l} \]
The time taken \( t_3 \) can be found using:
\[ V_m = a_3 t_3 \]
Thus:
\[ t_3 = \frac{V_m}{a_3} = \frac{V_m}{\frac{V_m^2}{6l}} = \frac{6l}{V_m} \]

### Step 4: Calculate total time and average speed
Now we can find the total time \( T \):
\[ T = t_1 + t_2 + t_3 = \sqrt{\frac{2l}{a_1}} + \frac{2l}{V_m} + \frac{6l}{V_m} \]
\[ T = \sqrt{\frac{2l}{a_1}} + \frac{8l}{V_m} \]

The total distance \( D \) traveled is:
\[ D = l + 2l + 3l = 6l \]

The average speed \( V_{avg} \) is:
\[ V_{avg} = \frac{D}{T} = \frac{6l}{\sqrt{\frac{2l}{a_1}} + \frac{8l}{V_m}} \]

### Step 5: Find the maximum speed
The maximum speed \( V_m \) occurs at the end of the first phase:
Using the equation:
\[ V_m^2 = 2a_1 l \implies V_m = \sqrt{2a_1 l} \]

### Step 6: Find the ratio of average speed to maximum speed
Now we need to find the ratio \( \frac{V_{avg}}{V_m} \):
\[ \frac{V_{avg}}{V_m} = \frac{6l}{\sqrt{\frac{2l}{a_1}} + \frac{8l}{V_m}} \cdot \frac{1}{\sqrt{2a_1 l}} \]
This simplifies to:
\[ = \frac{6}{\sqrt{\frac{2}{a_1}} + \frac{8}{\sqrt{2a_1 l}}} \]

After simplification, we find that:
\[ \frac{V_{avg}}{V_m} = \frac{3}{5} \]

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