For first \(5\, sec\) motion \({s_5} = 10\;metre\)

\(s = ut + \frac{1}{2}a{t^2} \Rightarrow 10 = 5u + \frac{1}{2}a{(5)^2}\)

\(2u + 5a = 4\) …(i)

For first \(8 \,sec\) of motion \({s_8} = 20\;metre\)

\(20 = 8u + \frac{1}{2}a{(8)^2} \Rightarrow 2u + 8a = 5\) …(ii)

By solving \(u = \frac{7}{6}m/s\;{\rm{and }}a = \frac{1}{3}m/{s^2}\)

Now distance travelled by particle in Total \(10\) sec.

\({s_{10}} = u \times 10 + \frac{1}{2}a{(10)^2}\)

By substituting the value of \(u\) and \(a\) we will get \({s_{10}} = 28.3\;m\)

so the distance in last \(2\;\sec = {s_{10}} - {s_8}\)

\( = 28.3 - 20 = 8.3\,m\)