A lift ascends with a constant acceleration of `4 m s^(-2)` then with a constant velocity v and finally stops under a constant retardation of `4 m s^(-2)`. If the total height ascended be 20 m and the total time taken is 6 s then the time during which the lift was movitng with a vclocity `nu` is

Options

A. 2a
B. 3s
C. 4s
D. 5s

Explanation

To solve the problem step-by-step, we will analyze the motion of the lift in three distinct phases: acceleration, constant velocity, and deceleration.

### Step 1: Define the Variables
Let:
- \( a = 4 \, \text{m/s}^2 \) (acceleration)
- \( u = 0 \, \text{m/s} \) (initial velocity)
- \( v \) = final velocity after acceleration
- \( t_1 \) = time taken to accelerate
- \( t \) = time taken to move with constant velocity
- \( t_2 \) = time taken to decelerate
- Total height ascended = 20 m
- Total time taken = 6 s

### Step 2: Relate the Times
Since the lift accelerates and decelerates with the same magnitude of acceleration, we have:
\[ t_2 = t_1 \]
Thus, the total time can be expressed as:
\[ t_1 + t + t_1 = 6 \]
or
\[ 2t_1 + t = 6 \]
(1)

### Step 3: Calculate the Final Velocity
Using the first equation of motion, the final velocity \( v \) after time \( t_1 \) is:
\[ v = u + at_1 = 0 + 4t_1 = 4t_1 \]
(2)

### Step 4: Calculate the Displacement
The total displacement can be calculated as the sum of displacements during each phase:
1. **During acceleration**:
\[ s_1 = ut_1 + \frac{1}{2} a t_1^2 = 0 + \frac{1}{2} (4) t_1^2 = 2t_1^2 \]
2. **During constant velocity**:
\[ s_2 = vt = (4t_1)t = 4t_1 t \]
3. **During deceleration**:
\[ s_3 = vt_2 - \frac{1}{2} a t_2^2 = (4t_1)t_1 - \frac{1}{2} (4) t_1^2 = 4t_1^2 - 2t_1^2 = 2t_1^2 \]

The total displacement is:
\[ s = s_1 + s_2 + s_3 = 2t_1^2 + 4t_1 t + 2t_1^2 = 4t_1^2 + 4t_1 t \]
Setting this equal to the total height ascended:
\[ 4t_1^2 + 4t_1 t = 20 \]
Dividing through by 4:
\[ t_1^2 + t_1 t = 5 \]
(3)

### Step 5: Substitute from Equation (1)
From equation (1), we can express \( t \) in terms of \( t_1 \):
\[ t = 6 - 2t_1 \]
Substituting this into equation (3):
\[ t_1^2 + t_1(6 - 2t_1) = 5 \]
Expanding this gives:
\[ t_1^2 + 6t_1 - 2t_1^2 = 5 \]
Combining like terms:
\[ -t_1^2 + 6t_1 - 5 = 0 \]
Multiplying through by -1:
\[ t_1^2 - 6t_1 + 5 = 0 \]

### Step 6: Solve the Quadratic Equation
Factoring the quadratic:
\[ (t_1 - 1)(t_1 - 5) = 0 \]
Thus, \( t_1 = 1 \) or \( t_1 = 5 \).

### Step 7: Determine Validity of Solutions
Since \( t_1 + t + t_1 = 6 \), if \( t_1 = 5 \), then \( 2t_1 = 10 \), which is not possible. Therefore, we take:
\[ t_1 = 1 \]

### Step 8: Find Time with Constant Velocity
Substituting \( t_1 = 1 \) back into equation (1):
\[ 2(1) + t = 6 \]
Thus,
\[ t = 6 - 2 = 4 \]

### Final Answer
The time during which the lift was moving with a constant velocity \( v \) is:
\[ \boxed{4 \, \text{s}} \]