d
(d)\(S = \frac{{{t^3}}}{3}\)
\(dS = {t^2}dt\)
\(a = \frac{{{d^2}S}}{{d{t^2}}} = \frac{{{d^2}}}{{d{t^2}}}\left[ {\frac{{{t^3}}}{3}} \right] = 2t\;m/{s^2}\)
Now work done by the force \(W = \int\limits_0^2 {F.dS} = \int\limits_0^2 {ma.dS} \)
\(\int\limits_0^2 {3 \times 2t \times {t^2}dt} \)\( = \int\limits_0^2 {6{t^3}dt} \)= \(\frac{3}{2}\left[ {{t^4}} \right]_0^2\)
\(= 24 \,J\)