$x=\frac{1}{2}a{{t}^{2}}+{{v}_{{}^\circ }}t+{{x}_{{}^\circ }}\xrightarrow[{{v}_{{}^\circ }} B=0]{{{v}_{{}^\circ }}A=0}\left\{ \begin{matrix} {{x}_{A}}=\frac{1}{2}{{a}_{A}}{{t}^{2}}+{{x}_{{}^\circ }}A  \\ {{x}_{B}}=\frac{1}{2}{{a}_{B}}{{t}^{2}}+{{x}_{{}^\circ }}B  \\ \end{matrix} \right.$

$\xrightarrow[{{x}_{A}}={{x}_{B}}]{t=2s}\frac{1}{2}{{a}_{A}}\times {{2}^{2}}+{{x}_{{}^\circ }}A=\frac{1}{2}{{a}_{B}}\times {{2}^{2}}+{{x}_{{}^\circ }}B$

$\xrightarrow{{{x}_{{}^\circ }}B-{{x}_{{}^\circ }}A=15m}2({{a}_{A}}-{{a}_{B}})=15$

$\Rightarrow {{a}_{A}}-{{a}_{B}}=\frac{15}{2}\frac{m}{{{s}^{2}}}\xrightarrow{v=at+{{v}_{{}^ \circ }},{{v}_{{}^\circ }}A={{v}_{{}^\circ }}B=0}\left\{ \begin{matrix} {{v}_{A}}={{a}_{A}}t  \\ {{v}_{B}}={{a}_{B}}t  \\ \end{matrix} \right.$

$\Rightarrow {{v}_{A}}-{{v}_{B}}=({{a}_{A}}-{{a}_{B}})t\xrightarrow[{{v}_{A}}-{{v}_{B}}=12\frac{m}{s}]{{{a}_{A}}-{{a}_{B}}=\frac{15}{2}\frac{m}{{{s}^{2}}}}12=\frac{15}{2}t$

$\Rightarrow t=\frac{24}{15}=\frac{8}{5}=1/6s$