An elevator car whose floor to ceiling distance is equal to `2.7m` starts ascending with constant acceleration `1.2 m//s^2.` 2 s after the start, a bolt begins falling from the ceiling of the car. Find
(a)the time after which bolt hits the floor of the elevator.
(b)the net displacement and distance travelled by the bolt, with respect to earth. (Take `g=9.8 m//s^2)`

Explanation

To solve the problem step by step, let's break it down into parts (a) and (b).

### Given Data:
- Height of the elevator car (h) = 2.7 m
- Acceleration of the elevator (a) = 1.2 m/s²
- Time after which the bolt falls (t₀) = 2 s
- Acceleration due to gravity (g) = 9.8 m/s²

### Part (a): Time after which the bolt hits the floor of the elevator

1. **Calculate the velocity of the elevator after 2 seconds:**
\[ v = u + at \]
Here, \(u = 0\) (initial velocity), \(a = 1.2 \, \text{m/s}^2\), and \(t = 2 \, \text{s}\).
\[ v = 0 + (1.2 \times 2) = 2.4 \, \text{m/s} \]

2. **Determine the relative acceleration of the bolt:**
When the bolt falls, it experiences two accelerations: the upward acceleration of the elevator (1.2 m/s²) and the downward acceleration due to gravity (9.8 m/s²). The net acceleration acting on the bolt is:
\[ a_{\text{net}} = g + a = 9.8 + 1.2 = 11 \, \text{m/s}^2 \]

3. **Use the equation of motion to find the time taken by the bolt to hit the floor:**
The distance the bolt needs to fall is equal to the height of the elevator, which is 2.7 m. Using the equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \(s = 2.7 \, \text{m}\), \(u = 0\) (the bolt starts from rest relative to the elevator), and \(a = 11 \, \text{m/s}^2\).
\[ 2.7 = 0 \cdot t + \frac{1}{2} \cdot 11 \cdot t^2 \]
Simplifying gives:
\[ 2.7 = \frac{11}{2} t^2 \quad \Rightarrow \quad t^2 = \frac{2 \cdot 2.7}{11} \quad \Rightarrow \quad t^2 = 0.4909 \]
\[ t = \sqrt{0.4909} \approx 0.7 \, \text{s} \]

### Part (b): Net displacement and distance travelled by the bolt with respect to the earth

1. **Calculate the total time from the start of the motion until the bolt hits the floor:**
The total time is the time before the bolt starts falling plus the time it takes to hit the floor:
\[ T = t_0 + t = 2 + 0.7 = 2.7 \, \text{s} \]

2. **Calculate the displacement of the bolt with respect to the ground:**
Using the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \(u = 2.4 \, \text{m/s}\) (initial velocity of the bolt), \(a = -9.8 \, \text{m/s}^2\) (acceleration due to gravity), and \(t = 0.7 \, \text{s}\):
\[ s = (2.4 \cdot 0.7) + \frac{1}{2} \cdot (-9.8) \cdot (0.7)^2 \]
\[ s = 1.68 - 2.401 = -0.721 \, \text{m} \]
This negative sign indicates that the bolt has moved downward relative to its initial position.

3. **Calculate the total distance travelled by the bolt:**
The bolt first moves upward and then downward. The upward distance until it stops is given by:
\[ s_1 = \frac{u^2}{2g} = \frac{(2.4)^2}{2 \cdot 9.8} = \frac{5.76}{19.6} \approx 0.293 \, \text{m} \]
The downward distance after it stops until it hits the floor is:
\[ s_2 = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot (0.7)^2 = \frac{1}{2} \cdot 9.8 \cdot 0.49 \approx 2.401 \, \text{m} \]
Therefore, the total distance travelled by the bolt is:
\[ \text{Total distance} = s_1 + s_2 \approx 0.293 + 2.401 \approx 2.694 \, \text{m} \]

### Summary of Results:
- (a) Time after which the bolt hits the floor: **0.7 seconds**
- (b) Net displacement with respect to the earth: **-0.721 m** (downward)
- Total distance travelled by the bolt: **2.694 m**