Question

In the figure shown, blocks $A$ and $C$ start from rest and move to the right with acceleration $a_{A}=12 \,t \,m / s ^{2}$ and $a_{C}=3\, m / s ^{2} .$ Here, $t$ is in seconds. The time (in $s$) when block $B$ again comes to rest is

Answer 1

$2 a_{B}=(3-12 t)$
$2\left(\frac{d v_{B}}{d t}\right)=(3-12 t)$
$2 \int\limits_{0}^{0} d v_{B}=3 t-12 \frac{t^{2}}{2}$
$0=t(3-6 t)$
$\Rightarrow t=\frac{1}{2} s$