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 segments: acceleration, constant velocity, and deceleration. We will denote the time spent in each segment as \( t_1 \) (for acceleration), \( t_2 \) (for constant velocity), and \( t_3 \) (for deceleration).

### Step 1: Understanding the Motion

1. **Acceleration Phase (0 to \( t_1 \))**:
- Initial velocity \( u = 0 \)
- Acceleration \( a = 1 \, \text{m/s}^2 \)
- Final velocity after acceleration \( v = u + at_1 = 0 + 1 \cdot t_1 = t_1 \)
- Distance covered during acceleration \( s_1 = ut_1 + \frac{1}{2} a t_1^2 = 0 + \frac{1}{2} \cdot 1 \cdot t_1^2 = \frac{1}{2} t_1^2 \)

2. **Constant Velocity Phase (\( t_1 \) to \( t_1 + t_2 \))**:
- Distance covered during this phase \( s_2 = v \cdot t_2 = (t_1) \cdot t_2 = t_1 t_2 \)

3. **Deceleration Phase (\( t_1 + t_2 \) to \( t_1 + t_2 + t_3 \))**:
- Initial velocity \( u = t_1 \)
- Final velocity \( v = 0 \)
- Retardation \( a = 1 \, \text{m/s}^2 \)
- Using the equation \( v^2 = u^2 - 2as \) to find the distance covered during deceleration:
\[ 0 = t_1^2 - 2 \cdot 1 \cdot s_3 \implies s_3 = \frac{t_1^2}{2} \]

### Step 2: Total Distance and Total Time

We know that the total distance ascended by the lift is 7 m and the total time for the journey is 8 s.

- Total distance:
\[ s_1 + s_2 + s_3 = \frac{1}{2} t_1^2 + t_1 t_2 + \frac{1}{2} t_1^2 = t_1^2 + t_1 t_2 = 7 \quad \text{(Equation 1)} \]

- Total time:
\[ t_1 + t_2 + t_3 = t_1 + t_2 + t_1 = 2t_1 + t_2 = 8 \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

From Equation 2, we can express \( t_2 \) in terms of \( t_1 \):
\[ t_2 = 8 - 2t_1 \quad \text{(Substituting in Equation 1)} \]

Now substituting \( t_2 \) into Equation 1:
\[ t_1^2 + t_1(8 - 2t_1) = 7 \]
\[ t_1^2 + 8t_1 - 2t_1^2 = 7 \]
\[ -t_1^2 + 8t_1 - 7 = 0 \]
\[ t_1^2 - 8t_1 + 7 = 0 \]

### Step 4: Solve the Quadratic Equation

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t_1 = \frac{8 \pm \sqrt{(8)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} \]
\[ t_1 = \frac{8 \pm \sqrt{64 - 28}}{2} \]
\[ t_1 = \frac{8 \pm \sqrt{36}}{2} \]
\[ t_1 = \frac{8 \pm 6}{2} \]
Thus, we have two possible values:
1. \( t_1 = \frac{14}{2} = 7 \)
2. \( t_1 = \frac{2}{2} = 1 \)

### Step 5: Calculate \( t_2 \)

Using \( t_1 = 1 \):
\[ t_2 = 8 - 2 \cdot 1 = 6 \]

Using \( t_1 = 7 \):
\[ t_2 = 8 - 2 \cdot 7 = -6 \quad \text{(not valid)} \]

### Conclusion

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