Let initial \(\left(t=0\right)\) velocity of particle = u

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

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

2u + 5a = 4 …(i)

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

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

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

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