and $x = \int_0^t {v\;dt} = \int_0^t {(4{t^3} - 2t)} \;dt = {t^4} - {t^2}$
When particle is at 2m from the origin ${t^4} - {t^2} = 2$
$⇒$ ${t^4} - {t^2} - 2 = 0$ $({t^2} - 2)\;({t^2} + 1) = 0$ $⇒$ $t = \sqrt 2 \;\sec $
Acceleration at $t = \sqrt {2\;} \;\sec $ given by,
$a = 12{t^2} - 2$$ = 12 \times 2 - 2$= $22\;m/{s^2}$
($g = 9.8\,m/{s^2}$)