The velocity$-$time graph of an ascending passenger lift is as shown in the figure below.
$(i)$ Identify the kind of motion of the lift represented by lines $OA$ and $BC$.
$(ii)$ Calculate the acceleration of the lift:
$(a)$ During the first two seconds.
$(b)$ Between the $3^{rd}$ and $10^{th}$ second.
$(c)$ During the last two seconds.

(N/A) $(i)$ The motion represented by line $OA$ is uniformly accelerated motion,and the motion represented by line $BC$ is uniformly retarded motion (deceleration).
$(ii)$ The acceleration of the lift is calculated as follows:
$(a)$ During the first two seconds (from $OA$):
Initial velocity $u = 0 \ m/s$,final velocity $v = 4.6 \ m/s$,time $t = 2 \ s$.
$a = \frac{v - u}{t} = \frac{4.6 - 0}{2} = 2.3 \ m/s^2$.
$(b)$ Between the $3^{rd}$ and $10^{th}$ second,the graph is a straight line parallel to the time axis,which indicates constant velocity. Therefore,the acceleration is $0 \ m/s^2$.
$(c)$ During the last two seconds (from $BC$):
Initial velocity $u = 4.6 \ m/s$,final velocity $v = 0 \ m/s$,time $t = 12 - 10 = 2 \ s$.
$a = \frac{v - u}{t} = \frac{0 - 4.6}{2} = -2.3 \ m/s^2$.