A body starts from rest & moves with constant acceleration `a_1` for time `t_1` then it retards uniformly with `a_2`, in time `t_2 ` and finally comes at rest. Find `(t_1)/(t_2)`

Answer

To solve the problem, we need to analyze the motion of the body in three phases: acceleration, uniform motion, and deceleration.

### Step-by-Step Solution:

1. **Phase 1: Acceleration**
- The body starts from rest and accelerates with acceleration \( a_1 \) for a time \( t_1 \).
- The initial velocity \( u = 0 \).
- The final velocity after this phase, \( V_{max} \), can be calculated using the equation of motion:
\[ V_{max} = u + a_1 t_1 = 0 + a_1 t_1 = a_1 t_1 \]
- Thus, we have:
\[ V_{max} = a_1 t_1 \quad \text{(Equation 1)} \]

2. **Phase 2: Deceleration**
- After \( t_1 \), the body starts to decelerate uniformly with acceleration \( a_2 \) for a time \( t_2 \).
- The initial velocity for this phase is \( V_{max} = a_1 t_1 \).
- The final velocity after this phase is \( 0 \) (the body comes to rest).
- Using the equation of motion for this phase:
\[ 0 = V_{max} - a_2 t_2 \]
- Substituting \( V_{max} \) from Equation 1:
\[ 0 = a_1 t_1 - a_2 t_2 \]
- Rearranging gives:
\[ a_1 t_1 = a_2 t_2 \quad \text{(Equation 2)} \]

3. **Finding the Ratio \( \frac{t_1}{t_2} \)**
- From Equation 2, we can express the ratio of \( t_1 \) to \( t_2 \):
\[ \frac{t_1}{t_2} = \frac{a_2}{a_1} \]

### Final Answer:
\[ \frac{t_1}{t_2} = \frac{a_2}{a_1} \]