The bus's position [tex]x[/tex] at time [tex]t[/tex] is
[tex]x_{\rm bus}=\dfrac12\left(2\,\dfrac{\rm m}{\rm s^2}\right)t^2[/tex]
If we take the bus's rear's starting position to be the origin, then the position of the mate (?) is given by
[tex]x_{\rm mate}=-9\,\mathrm m+\left(6\,\dfrac{\rm m}{\rm s}\right)t[/tex]
The mate draws level with the rear of the bus when [tex]x_{\rm bus}=x_{\rm mate}[/tex]:
[tex]\dfrac12\left(2\,\dfrac{\rm m}{\rm s^2}\right)t^2=-9\,\mathrm m+\left(6\,\dfrac{\rm m}{\rm s}\right)t[/tex]
Drop the units to make things simpler:
[tex]\dfrac22t^2=-9+6t\iff t^2-6t+9=(t-3)^2=0\implies t=3[/tex]
So it would take 3 seconds for the mate to catch up to the bus.
