A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation `2a.` If t is the time of ascent, find the depth of the shaft.

Options

A. `(at^2)/4`
B. `(at^2)/3`
C. `(at^2)/2`
D. `(at^2)/8`

Explanation

To solve the problem of finding the depth of the shaft where a lift ascends with uniform acceleration and then with uniform retardation, we can follow these steps:

### Step 1: Define the time intervals
Let \( t_0 \) be the time during which the lift accelerates with acceleration \( a \). The time during which the lift decelerates (retards) with retardation \( 2a \) will be \( \frac{t_0}{2} \) because the retardation is double the acceleration.

### Step 2: Relate the total time to the time intervals
The total time of ascent \( t \) is the sum of the time of acceleration and the time of retardation:
\[ t = t_0 + \frac{t_0}{2} \]
This simplifies to:
\[ t = \frac{3t_0}{2} \]
From this, we can solve for \( t_0 \):
\[ t_0 = \frac{2t}{3} \]

### Step 3: Calculate the distance during acceleration
The distance \( S_1 \) traveled during the acceleration phase can be calculated using the formula:
\[ S_1 = ut + \frac{1}{2} a t_0^2 \]
Since the initial velocity \( u = 0 \) (the lift starts from rest), this simplifies to:
\[ S_1 = \frac{1}{2} a t_0^2 \]
Substituting \( t_0 = \frac{2t}{3} \):
\[ S_1 = \frac{1}{2} a \left(\frac{2t}{3}\right)^2 = \frac{1}{2} a \cdot \frac{4t^2}{9} = \frac{2at^2}{9} \]

### Step 4: Calculate the distance during retardation
The distance \( S_2 \) traveled during the retardation phase can be calculated using the formula:
\[ S_2 = vt - \frac{1}{2} a t^2 \]
To find \( v \) (the final velocity at the end of the acceleration phase), we use:
\[ v = u + at_0 = 0 + a \left(\frac{2t}{3}\right) = \frac{2at}{3} \]
Now substituting \( v \) into the formula for \( S_2 \):
\[ S_2 = \left(\frac{2at}{3}\right) \left(\frac{t}{2}\right) - \frac{1}{2} (2a) \left(\frac{t_0}{2}\right)^2 \]
Calculating \( S_2 \):
\[ S_2 = \frac{2at^2}{6} - \frac{1}{2} (2a) \left(\frac{t}{3}\right)^2 = \frac{2at^2}{6} - \frac{1}{2} (2a) \cdot \frac{t^2}{9} = \frac{2at^2}{6} - \frac{at^2}{9} \]
Finding a common denominator (18):
\[ S_2 = \frac{6at^2}{18} - \frac{2at^2}{18} = \frac{4at^2}{18} = \frac{2at^2}{9} \]

### Step 5: Calculate the total distance (depth of the shaft)
The total distance \( S \) (depth of the shaft) is the sum of \( S_1 \) and \( S_2 \):
\[ S = S_1 + S_2 = \frac{2at^2}{9} + \frac{2at^2}{9} = \frac{4at^2}{9} \]

### Final Answer
Thus, the depth of the shaft is:
\[ S = \frac{4at^2}{9} \]