Why different methods to solve this question gives different values of time While solving the following question.Why do we get different values of time by different approaches.
An elevator whose floor to ceiling distance is $2.7m$  starts ascending with a constant acceleration of $1.2m/s^{2}$.Two sec after it starts, a bolt begins to fall from the ceiling of the elevator.Find the bolt's free fall time.
Approach 1 :- In lift's frame of reference.
Applying $S=ut+\frac12 at^2$ on bolt we get,
$2.7=0+\frac12 (9.8)t^2$
$t=0.742 sec$
Approach 2:- In ground frame of reference.
Let bolt takes ‘t’ time to fall
$\vec{s_{b,l}}=-2.7\vec j$
$\vec{s_{l,g}}=\frac {1.2}{2}((t+2)^2-4) \vec j$
As $\vec{s_{b,l}}+\vec{s_{l,g}}=\vec{s_{b,g}}$
So, $\vec{s_{b,g}}=(\frac {1.2}{2}((t+2)^2-4)-2.7)\vec j$
Velocity  of bolt after $2$ sec $=2.4m/s$
Applying $S=ut+\frac12 at^2$ on bolt we get,
$\frac {1.2}{2}((t+2)^2-4)-2.7=2.4t-\frac12 (9.8)t^2$
On solving $t=0.7006$ sec
 A: In approach 1 which is from the lift frame, you are doing the calculation as if the frame is inertial. Meaning, you are using only the gravitational acceleration $g$ and are ignoring the centrifugal part, namely the presence of the lift acceleration.
$$\text{Inertial frame:}\qquad a=g$$
This is how the bolt will fall if dropped here on Earth. This is how it would fall if the lift wasn't accelerating. But the lift is accelerating up also. The bolt will reach the floor quicker since the floor is accelerating upwards towards the bolt. The lift acceleratino must be added so that you'll use an adjusted frame-specific acceleration in the formula:
$$\text{Non-inertial lift frame:} \qquad a=g+a_\text{lift}$$
In general you must always add (or remove) the centrifugal acceleration part when using a non-inertial frame. This is the part that distinguishes inertial from non-inertial in our calculations. Then the calculation fits:
$$S=ut+\frac12 \underbrace{(g+a_\text{lift})}_a t^2\quad \Leftrightarrow \quad t=0.7006\,\mathrm s$$
