A lift ascends with constant acceleration `a=1ms^(-2)`, then with constant velocity and finally, it stops under constant retardation `a=1ms^(-2)`. If total distance ascended by the lift is 7 m, in a total time of the journey is 8 s. Find the time (in second) for which lift moves with constant velocity.

Answer

To solve the problem, we need to analyze the motion of the lift in three phases:

1. **Phase 1**: The lift ascends with constant acceleration \( a = 1 \, \text{m/s}^2 \).
2. **Phase 2**: The lift moves with constant velocity.
3. **Phase 3**: The lift stops under constant retardation \( a = 1 \, \text{m/s}^2 \).

Let:
- \( t_1 \) = time during acceleration
- \( t_2 \) = time during constant velocity
- \( t_3 \) = time during retardation

### Step 1: Set up equations for time
From the problem, we know that the total time of the journey is:
\[ t_1 + t_2 + t_3 = 8 \, \text{s} \quad \text{(1)} \]

### Step 2: Set up equations for distance
The total distance ascended by the lift is 7 m. We can express the distance traveled in each phase.

- **Distance during acceleration (Phase 1)**:
Using the formula for distance under constant acceleration:
\[ S_1 = ut + \frac{1}{2} a t_1^2 \]
Since the initial velocity \( u = 0 \):
\[ S_1 = \frac{1}{2} \cdot 1 \cdot t_1^2 = \frac{1}{2} t_1^2 \quad \text{(2)} \]

- **Distance during constant velocity (Phase 2)**:
The velocity at the end of Phase 1 is:
\[ v = u + at = 0 + 1 \cdot t_1 = t_1 \]
Thus, the distance during this phase is:
\[ S_2 = v \cdot t_2 = t_1 \cdot t_2 \quad \text{(3)} \]

- **Distance during retardation (Phase 3)**:
Using the formula for distance during constant retardation:
\[ S_3 = vt - \frac{1}{2} a t_3^2 \]
Here, the initial velocity is \( v = t_1 \) and \( a = 1 \):
\[ S_3 = t_1 t_3 - \frac{1}{2} \cdot 1 \cdot t_3^2 = t_1 t_3 - \frac{1}{2} t_3^2 \quad \text{(4)} \]

### Step 3: Set up the total distance equation
The total distance is:
\[ S_1 + S_2 + S_3 = 7 \quad \text{(5)} \]
Substituting equations (2), (3), and (4) into (5):
\[ \frac{1}{2} t_1^2 + t_1 t_2 + \left(t_1 t_3 - \frac{1}{2} t_3^2\right) = 7 \]
This simplifies to:
\[ \frac{1}{2} t_1^2 + t_1 t_2 + t_1 t_3 - \frac{1}{2} t_3^2 = 7 \quad \text{(6)} \]

### Step 4: Substitute \( t_3 \)
From equation (1), we can express \( t_3 \) in terms of \( t_1 \) and \( t_2 \):
\[ t_3 = 8 - t_1 - t_2 \quad \text{(7)} \]
Substituting (7) into (6):
\[ \frac{1}{2} t_1^2 + t_1 t_2 + t_1 (8 - t_1 - t_2) - \frac{1}{2} (8 - t_1 - t_2)^2 = 7 \]

### Step 5: Simplify and solve for \( t_1 \) and \( t_2 \)
Expanding and simplifying this equation will yield a quadratic equation in terms of \( t_1 \) and \( t_2 \). After simplification, we can solve for \( t_1 \) and then use it to find \( t_2 \).

After solving the quadratic equation, we find:
- \( t_1 = 1 \, \text{s} \)
- Substitute \( t_1 \) back into equation (1) to find \( t_2 \):
\[ 1 + t_2 + t_3 = 8 \implies t_2 + t_3 = 7 \]

Since \( t_3 = t_1 = 1 \, \text{s} \):
\[ t_2 = 7 - 1 - 1 = 6 \, \text{s} \]

### Final Answer
The time for which the lift moves with constant velocity is:
\[ \boxed{6 \, \text{s}} \]