1D Kinematics - Relative Motion involving non-inertial frames of reference I recently came across a question involving non-inertial frames of reference. I didn't quite understand the way it had been solved due to some conceptual confusion regarding certain deductions made to solve the problem.
The question is this:

QUESTION:
  An elevator is ascending with an acceleration $0.2 \ m s^{-2}$. At the instant the elevator's velocity is $1 \ m s^{-1}$, a loose bolt on the elevator's ceiling drops down. The height between the ceiling and floor of the elevator is $5 \ m$. Find out the time taken for the bolt to reach the floor and the distance it covered. 

MY ANSWER:
Since with respect to the ground, the initial velocities of both the elevator and the bolt are $1 \ m s^{-1}$ at that particular instant, the bolt's velocity relative to the elevator is $0 \ m s^{-1}$. The relative displacement of the bolt with respect to the elevator floor will be $5 \ m$  after the time taken for the bolt to reach the floor. Since the elevator is an accelerating frame of reference, there will be a fictitious (pseudo) force acting in the opposite direction to the acceleration of the elevator but equal in magnitude. So not only does the bolt undergo free fall due to acceleration due to gravity (which I have taken as $9.8 \ m s^{-2}$), but also experiences an additional acceleration due to this pseudo-force. If I have considered the direction the elevator is moving in to be in the positive direction, then the total downward acceleration of the bolt is $= -9.8 + (-0.2) = -10 \ m s^{-2}$. But I need to consider the bolt's relative acceleration with respect to the floor of the elevator $= (-10 + 0.2) = -9.8 \ m s^{-2}$. This is as far as I can come. When I try substituting this into an equation, mathematical trouble brews and I am unable to solve the problem.
Please enlighten as to whether my ideas are correct and if not how would one go about solving this particular question using the ideas of relative motion?
 A: Let us assume we are standing inside the lift. For us, relative initial velocity $(u_{rel})$ is $0\:ms^{-1}$. The relative diplacement $s_{rel}$ is $5\:m$. The relative acceleration $a_{rel}$ is $10\:ms^{-2}$ (since the lift is going up with an acceleration of $0.2ms^{-2}$ and the bolt is falling with an acceleration of $9.8ms^{-2}$, it would appear to be accelerating towards us with $10ms^{-2}$. You can verify with the formula $a_{bolt,\;lift}=a_{bolt}-a_{lift}$, where $a_{bolt, \;lift}$ is the acceleration of $bolt$ with respect to $lift$, and acceleration of lift is $-0.2ms^{-2}$ while that of bolt is $+9.8ms^{-2}$) Putting this data in the kinematical equation $s_{rel}=u_{rel}+\frac12a_{rel}t^2$, we get $t=1s$
Now we are standing outside the lift. Now, initial velocity of the bolt is $u=-1ms^{-1}$, acceleration of the bolt is $a=+9.8ms^{-2}$. Time is $t=1s$. Again putting this data in the equation $s=ut+\frac12at^2$, we get $s=3.9m$.
Thus, the bolt covers a distance of $3.9m$ in $1s$. We only have to be careful about the sign of the physical quantities in relative motion.
A: Taking $g = 9.8m/s^2$
Then $g_{effective} = 9.8 + 0.2 = 10 m/s^2$
where $g_{eff}$ is the acceleration of the bolt with respect to the elevator and is directed downwards.
Now $s = 5m,u = 1m/s,a = -10m/s^2$(directed downwards)
Substitute these into $s = ut + 0.5at^2$ and solve for time.
