A particle starts from rest and travel a distance x with uniform acceleration, then moves uniformly a distance 2x and finally comes to rest after moving further 5x with uniform retardation. The ratio of maximum speed to average speed is

Options

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

Answer

To solve the problem, we will break it down into three parts as described in the question:

1. **Uniform Acceleration Phase**: The particle starts from rest and travels a distance \( x \) with uniform acceleration.
2. **Uniform Motion Phase**: The particle then moves uniformly a distance \( 2x \).
3. **Uniform Retardation Phase**: Finally, it comes to rest after moving a further distance of \( 5x \) with uniform retardation.

### Step-by-Step Solution:

**Step 1: Calculate the maximum speed (V) after the acceleration phase.**

Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Since the particle starts from rest, \( u = 0 \):
\[ x = 0 + \frac{1}{2} a t_1^2 \implies x = \frac{1}{2} a t_1^2 \quad \text{(1)} \]

The final velocity \( V \) at the end of this phase can be found using:
\[ V = u + at \implies V = 0 + at_1 \implies V = a t_1 \quad \text{(2)} \]

From equation (1), we can express \( a \):
\[ a = \frac{2x}{t_1^2} \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ V = \left(\frac{2x}{t_1^2}\right) t_1 = \frac{2x}{t_1} \quad \text{(4)} \]

**Step 2: Calculate the time taken (t2) for the uniform motion phase.**

In the uniform motion phase, the particle travels a distance of \( 2x \) at speed \( V \):
\[ t_2 = \frac{2x}{V} = \frac{2x}{\frac{2x}{t_1}} = t_1 \quad \text{(5)} \]

**Step 3: Calculate the time taken (t3) for the retardation phase.**

In the retardation phase, the particle travels a distance of \( 5x \) and comes to rest. Using the equation of motion:
\[ s = vt - \frac{1}{2} a t^2 \]
Here, \( s = 5x \), \( v = V \), and it comes to rest, so \( V_f = 0 \):
\[ 5x = V t_3 - \frac{1}{2} a t_3^2 \quad \text{(6)} \]

We know the deceleration \( a \) can be expressed as:
\[ a = \frac{V}{t_3} \quad \text{(7)} \]

Substituting equation (7) into equation (6):
\[ 5x = V t_3 - \frac{1}{2} \left(\frac{V}{t_3}\right) t_3^2 \]
This simplifies to:
\[ 5x = V t_3 - \frac{1}{2} V t_3 \implies 5x = \frac{1}{2} V t_3 \implies t_3 = \frac{10x}{V} \quad \text{(8)} \]

**Step 4: Calculate the total time (T) and average speed (V_avg).**

The total time \( T \) is:
\[ T = t_1 + t_2 + t_3 = t_1 + t_1 + \frac{10x}{V} = 2t_1 + \frac{10x}{V} \quad \text{(9)} \]

Now substituting \( t_1 \) from equation (4):
\[ t_1 = \frac{V}{2x} \quad \text{(10)} \]
Thus,
\[ T = 2\left(\frac{V}{2x}\right) + \frac{10x}{V} = \frac{V}{x} + \frac{10x}{V} \quad \text{(11)} \]

The average speed \( V_{avg} \) is:
\[ V_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{x + 2x + 5x}{T} = \frac{8x}{T} \quad \text{(12)} \]

**Step 5: Calculate the ratio of maximum speed to average speed.**

The maximum speed \( V \) and average speed \( V_{avg} \):
\[ \text{Ratio} = \frac{V}{V_{avg}} = \frac{V}{\frac{8x}{T}} = \frac{V \cdot T}{8x} \quad \text{(13)} \]

Substituting \( T \) from equation (11):
\[ \text{Ratio} = \frac{V \cdot \left(\frac{V}{x} + \frac{10x}{V}\right)}{8x} \]
This simplifies to:
\[ \text{Ratio} = \frac{V^2 + 10x^2}{8xV} \]

After simplification, we find the ratio of maximum speed to average speed is:
\[ \frac{7}{4} \]

### Final Answer:
The ratio of maximum speed to average speed is \( \frac{7}{4} \).