Let initial r = D velocity of particle .

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

\(5 - \left(a + \frac{1}{2} a^{2}\right) = 10 - 5a + \frac{1}{2} a (5)^{2}\)

\(2u + 5\sigma = 4\) … (i)

For first 8 sec of motion \(s_a \quad 20 \quad metre\)

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

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

Now distance travelled by particle in Total 10 sec.

\(s_v = v \times 10 + \frac{1}{2} a 10^2\)

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

so the distance in last \(2 \sec \quad s_j - s_i\)

= 28.3 - 20 = 8.3 m