Starting from rest a particle is first accelerated for time `t_1` with constant acceleration `a_1` and then stops in time `t_2` with constant retardation `a_2.` Let `v_1` be the average velocity in this case and `s_1` the total displacement. In the second case it is accelerating for the same time `t_1` with constant acceleration `2a_1` and come to rest with constant retardation `a_2` in time `t_3.` If `v_2` is the average velocity in this case and `s_2` the total displacement, then

Options

A. `v_2=2v_1`
B. `2v_1ltv_2lt4v_1`
C. `s_2=2s_1`
D. `2s_1lts_2lt4s_1`

Explanation

To solve the problem, we need to analyze the motion of a particle under two different scenarios. Here’s a step-by-step breakdown of the solution:

### Step 1: Analyze the First Case
1. **Acceleration Phase**: The particle starts from rest and accelerates with constant acceleration \( a_1 \) for time \( t_1 \).
- Final velocity after acceleration, \( v_{\text{max}} = a_1 t_1 \).

2. **Retardation Phase**: The particle then comes to rest with constant retardation \( a_2 \) in time \( t_2 \).
- Using the equation \( v = u + at \), where \( u = v_{\text{max}} \) and \( v = 0 \):
\[ 0 = a_1 t_1 - a_2 t_2 \implies a_1 t_1 = a_2 t_2 \quad \text{(Equation 1)} \]

3. **Displacement Calculation**: The total displacement \( s_1 \) can be calculated as the area under the velocity-time graph.
- The area of the triangle during acceleration plus the area of the triangle during retardation:
\[ s_1 = \frac{1}{2} \times t_1 \times v_{\text{max}} + \frac{1}{2} \times t_2 \times v_{\text{max}} = \frac{1}{2} v_{\text{max}} (t_1 + t_2) \]
- Substitute \( v_{\text{max}} = a_1 t_1 \):
\[ s_1 = \frac{1}{2} a_1 t_1 (t_1 + t_2) \]

4. **Average Velocity Calculation**: The average velocity \( v_1 \) is given by:
\[ v_1 = \frac{s_1}{t_1 + t_2} = \frac{\frac{1}{2} a_1 t_1 (t_1 + t_2)}{t_1 + t_2} = \frac{1}{2} a_1 t_1 \]

### Step 2: Analyze the Second Case
1. **Acceleration Phase**: The particle accelerates with constant acceleration \( 2a_1 \) for the same time \( t_1 \).
- Final velocity after acceleration, \( v_{\text{max}} = 2a_1 t_1 \).

2. **Retardation Phase**: The particle comes to rest with constant retardation \( a_2 \) in time \( t_3 \).
- Using the equation \( 0 = 2a_1 t_1 - a_2 t_3 \):
\[ 2a_1 t_1 = a_2 t_3 \quad \text{(Equation 2)} \]

3. **Displacement Calculation**: The total displacement \( s_2 \) can be calculated similarly:
\[ s_2 = \frac{1}{2} \times t_1 \times v_{\text{max}} + \frac{1}{2} \times t_3 \times v_{\text{max}} = \frac{1}{2} v_{\text{max}} (t_1 + t_3) \]
- Substitute \( v_{\text{max}} = 2a_1 t_1 \):
\[ s_2 = \frac{1}{2} \times 2a_1 t_1 (t_1 + t_3) = a_1 t_1 (t_1 + t_3) \]

4. **Finding \( t_3 \)**: From Equation 2, we can express \( t_3 \):
- Rearranging gives:
\[ t_3 = \frac{2a_1 t_1}{a_2} \]

5. **Average Velocity Calculation**: The average velocity \( v_2 \) is given by:
\[ v_2 = \frac{s_2}{t_1 + t_3} = \frac{a_1 t_1 (t_1 + t_3)}{t_1 + t_3} \]

### Step 3: Comparing Average Velocities and Displacements
1. From the expressions for \( s_1 \) and \( s_2 \), we can compare:
\[ s_2 = 2 s_1 \quad \text{(since \( t_3 = 2t_2 \))} \]
Thus, \( s_2 > s_1 \).

2. For average velocities:
\[ v_2 = 2 v_1 \]
Thus, \( v_2 > v_1 \).

### Conclusion
- The relationships derived indicate that:
- \( v_2 > v_1 \)
- \( s_2 > s_1 \)