b
(b) \(v = 4{t^3} - 2t\) (given) \(\therefore \) \(a = \frac{{dv}}{{dt}} = 12{t^2} - 2\)

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,

\(a = 12{t^2} - 2\)\( = 12 \times 2 - 2\)= \(22\;m/{s^2}\)