A car, starting from rest, has a constant acceleration for a time interval ` t_1` during which it covers a distance In the next time interval the car has a constant retardation and comes to rest after covering a distance in time Which of the following relations is correct?

Options

A. ` (a_1) /(a_2) =(s_1)/( S_2) =(t_1)/(t_2) `
B. ` (a_1)/(a_2) =(s_2)/(s_1) =(t_1)/(t_2)`
C. ` (a_1)/(a_2) =(s_1)/(s_2) =(t_2)/(t_1)`
D. ` (a_1)/(a_2)=(s_2)/(s_1) =(t_2)/(t_1)`

Explanation

To solve the problem, we need to analyze the motion of the car during the two time intervals: the acceleration phase and the deceleration phase.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The car starts from rest, meaning its initial velocity \( u = 0 \).
- During the first time interval \( t_1 \), the car accelerates with a constant acceleration \( a_1 \) and covers a distance \( s_1 \).
- During the second time interval \( t_2 \), the car decelerates with a constant retardation \( a_2 \) and comes to rest after covering a distance \( s_2 \).

2. **Using Kinematic Equations**:
- For the first interval (acceleration):
\[ s_1 = ut + \frac{1}{2} a_1 t_1^2 \]
Since \( u = 0 \):
\[ s_1 = \frac{1}{2} a_1 t_1^2 \quad \text{(1)} \]

- For the second interval (deceleration):
\[ s_2 = v t_2 - \frac{1}{2} a_2 t_2^2 \]
Here, \( v \) is the final velocity after the first interval, which can be expressed as:
\[ v = a_1 t_1 \quad \text{(2)} \]
Substituting \( v \) into the equation for \( s_2 \):
\[ s_2 = (a_1 t_1) t_2 - \frac{1}{2} a_2 t_2^2 \quad \text{(3)} \]

3. **Relating the Accelerations**:
- The car comes to rest after the second interval, so the final velocity is zero:
\[ 0 = v - a_2 t_2 \]
Rearranging gives:
\[ a_2 = \frac{v}{t_2} = \frac{a_1 t_1}{t_2} \quad \text{(4)} \]

4. **Substituting \( a_2 \) into Equation (3)**:
- Substitute \( a_2 \) from equation (4) into equation (3):
\[ s_2 = (a_1 t_1) t_2 - \frac{1}{2} \left(\frac{a_1 t_1}{t_2}\right) t_2^2 \]
Simplifying this gives:
\[ s_2 = a_1 t_1 t_2 - \frac{1}{2} a_1 t_1 t_2 = \frac{1}{2} a_1 t_1 t_2 \quad \text{(5)} \]

5. **Finding the Relation**:
- From equations (1) and (5):
\[ s_1 = \frac{1}{2} a_1 t_1^2 \quad \text{and} \quad s_2 = \frac{1}{2} a_1 t_1 t_2 \]
- Taking the ratio \( \frac{s_1}{s_2} \):
\[ \frac{s_1}{s_2} = \frac{\frac{1}{2} a_1 t_1^2}{\frac{1}{2} a_1 t_1 t_2} = \frac{t_1}{t_2} \]
- Therefore, we have:
\[ \frac{s_1}{s_2} = \frac{t_1}{t_2} \]

6. **Final Relation**:
- We also know from the acceleration and deceleration that:
\[ \frac{a_1}{a_2} = \frac{t_2}{t_1} \]
- Thus, we conclude that:
\[ \frac{a_1}{a_2} = \frac{t_2}{t_1} = \frac{s_2}{s_1} \]

### Conclusion:
The correct relation is:
\[ \frac{a_1}{a_2} = \frac{t_2}{t_1} = \frac{s_2}{s_1} \]