a) Velocity vector is the derivative of position vector.

So to find velocity vector, differentiate position vector with respect to t. We get \(\displaystyle{v}{\left({t}\right)}={\left\langle\frac{{1}}{{5}},-\frac{{2}}{{t}^{{3}}},{4}{t}^{{3}}\right\rangle}\)

Now speed is the magnitude of velocity vector. That is,

Speed \(\displaystyle={\left|{\left|{v}\right|}\right|}\)

\(\displaystyle=\sqrt{{{\left(\frac{{1}}{{t}}\right)}^{{2}}+{\left(-\frac{{2}}{{t}^{{3}}}\right)}^{{2}}+{\left({4}{t}^{{3}}\right)}^{{2}}}}\)

\(\displaystyle=\sqrt{{\frac{{1}}{{t}^{{2}}}+\frac{{4}}{{t}^{{6}}}+{16}{t}^{{6}}}}\)

\(\displaystyle=\sqrt{{\frac{{{t}^{{4}}+{4}+{16}{t}^{{12}}}}{{t}^{{6}}}}}\)

\(\displaystyle=\frac{\sqrt{{{\left({16}{t}^{{12}}+{t}^{{4}}+{4}\right)}}}}{{t}^{{3}}}\)

Acceleration vector is the derivative of the velocity vector.

So to find acceleration vector, differentiate velocity vector with respect to t. We get,

\(a(t)=\left\langle-\frac{1}{t^{2}},\ -2(-\frac{3}{t^{4}}),\ 4(3t^{2}) \right\rangle\)

\(=\left\langle -\frac{1}{t^{2}},\ \frac{6}{t^{4}},\ 12t^{2}\right\rangle\)

b) Put \(\displaystyle{t}=\sqrt{{3}}\) in the velocity vector, we get

\(\displaystyle{v}{\left(\sqrt{{3}}\right)}={\left\langle\frac{{1}}{\sqrt{{3}}},-\frac{{2}}{{\left(\sqrt{{3}}\right)}^{{3}}},{4}{\left(\sqrt{{3}}\right)}^{{3}}\right\rangle}\)

\(\displaystyle={\left\langle\frac{{1}}{\sqrt{{3}}},-\frac{{2}}{{{3}\sqrt{{3}}}},{12}\sqrt{{3}}\right\rangle}\)

Put \(\displaystyle{t}=\sqrt{{3}}\) in the acceleration vector, we get

\(\displaystyle{a}{\left(\sqrt{{3}}\right)}={\left\langle-\frac{{1}}{{\left(\sqrt{{3}}\right)}^{{2}}},\frac{{6}}{{\left(\sqrt{{3}}\right)}^{{4}}},{12}{\left(\sqrt{{3}}\right)}^{{2}}\right\rangle}\)

\(\displaystyle={\left\langle-\frac{{1}}{{3}},\frac{{6}}{{9}},{12}{\left({3}\right)}\right\rangle}\)

\(\displaystyle={\left\langle-\frac{{1}}{{3}},\frac{{2}}{{3}},{36}\right\rangle}\)