Elevator In Shanghai Tower

Note that $a(t)=v'(t)$ . Since $a(t) > 0$ for $0 \le t < 3$ (I approximated the bounds based on the picture), we have that $v(t)$ is increasing on $0 \le t < 3$ . Since $a'(t)=0$ on $3 \le t \le 6$ , then $v(t)$ is constant for $3 \le t \le 6$ . Lastly, $a(t) < 0$ for $6 < t \le 9$ hence $v(t)$ is decreasing on $6 < t \le 9$ . We also know that $v(0)=0$ (elevator starts from rest) and that $\int_{0}^{9} a(t) \ dt = 0$ , hence the final velocity is $0 \frac{m}{s}$ , i.e $v(9)=0$ . This automatically rules out choice $B.)$ . Now from the remaining options, we observe that choices $C.)$ and $D.)$ both have corners which implies that $a(t)$ should be discontinuous on $0 \le t \le 9$ but it clearly is not. Hence the answer is $\boxed{A}$ , since it is the only that satisfies the conditions posed by the graph of $a(t)$ .

So basically acceleration is dv/dt. so acceleration = integration acc. over time. If we acceleration graph then acceleration is increasing or decreasing with time linearly at some times.So if we integrate it graph will be parabolic for sure at some times.So it rules out C and D.In B velocity is always increasing or constant but it can't be the case as in acceleration graph we can see that around 6 acceleration is increasing in negative that means velocity must be decreasing.