Answer: (a)(i) \(26\,\mathrm{m\,s^{-1}}\); (a)(ii) \(20\text{ m}\); (b) \(\displaystyle s=18\mathrm e^{3-t}+44t-130\).

Answer: (a)(i) \(26\,\mathrm{m\,s^{-1}}\); (a)(ii) \(20\text{ m}\); (b) \(\displaystyle s=18\mathrm e^{3-t}+44t-130\).

For \(0\leq t\leq3\),

\(a=6t.\)

Integrate to find the velocity:

\(v=3t^2+C.\)

When \(t=1\), \(v=2\), so

\(2=3(1)^2+C.\)

Thus \(C=-1\), and

\(v=3t^2-1.\)

(a)(i) At \(t=3\),

\(v=3(3)^2-1=27-1=26.\)

So the velocity is \(26\,\mathrm{m\,s^{-1}}\).

(a)(ii) Integrate \(v=3t^2-1\) to find displacement:

\(s=t^3-t+D.\)

When \(t=1\), \(s=-4\), so

\(-4=1-1+D.\)

Thus \(D=-4\), and

\(s=t^3-t-4.\)

At \(t=3\),

\(s=27-3-4=20.\)

(b) For \(t\geq3\),

\(a=18\mathrm e^{3-t}.\)

Integrate to find velocity:

\(v=-18\mathrm e^{3-t}+K.\)

The velocity must be continuous at \(t=3\), so \(v=26\) when \(t=3\):

\(26=-18\mathrm e^0+K=-18+K.\)

Hence \(K=44\), and

\(v=44-18\mathrm e^{3-t}.\)

Integrate again:

\(s=44t+18\mathrm e^{3-t}+L.\)

The displacement is \(20\) when \(t=3\), so

\(20=44(3)+18\mathrm e^0+L.\)

Thus

\(20=132+18+L,\)

so \(L=-130\). Therefore, for \(t\geq3\),

\(s=18\mathrm e^{3-t}+44t-130.\)