Answer: (a)(i) \(v=10\). (a)(ii) \(s=4\). (b) \(s=\frac{1569}{4}-\frac54\mathrm{e}^{-12}\approx392\).

For \(0\leqslant t\leqslant4\),

\(a=t^2-2.\)

Integrating,

\(v=\frac{t^3}{3}-2t+C.\)

Given \(v=-\frac13\) when \(t=3\),

\(-\frac13=9-6+C,\)

so \(C=-\frac{10}{3}\). Therefore

\(v=\frac{t^3}{3}-2t-\frac{10}{3}.\)

At \(t=4\),

\(v=\frac{64}{3}-8-\frac{10}{3}=10.\)

Now integrate \(v\):

\(s=\frac{t^4}{12}-t^2-\frac{10}{3}t+D.\)

Given \(s=-\frac14\) when \(t=3\),

\(-\frac14=\frac{81}{12}-9-10+D,\)

so \(D=12\). Hence

\(s=\frac{t^4}{12}-t^2-\frac{10}{3}t+12.\)

At \(t=4\),

\(s=4.\)

For \(4\leqslant t\leqslant10\),

\(a=19-5\mathrm{e}^{8-2t}.\)

Integrating,

\(v=19t+\frac52\mathrm{e}^{8-2t}+C.\)

Using \(v=10\) when \(t=4\),

\(10=76+\frac52+C,\)

so \(C=-\frac{137}{2}\). Therefore

\(v=19t+\frac52\mathrm{e}^{8-2t}-\frac{137}{2}.\)

Integrating again,

\(s=\frac{19t^2}{2}-\frac54\mathrm{e}^{8-2t}-\frac{137}{2}t+D.\)

Using \(s=4\) when \(t=4\), \(D=\frac{509}{4}\). Hence

\(s=\frac{19t^2}{2}-\frac54\mathrm{e}^{8-2t}-\frac{137}{2}t+\frac{509}{4}.\)

At \(t=10\),

\(s=\frac{1569}{4}-\frac54\mathrm{e}^{-12}=392.249\ldots.\)