Explanation

(a) Let initial $(t = 0)$ velocity of particle$= u$
For first $5\ sec$ of motion $s_{5} = 10\ metre$
$s = ut + \frac{1}{2}at^{2} \Rightarrow 10 = 5u + \frac{1}{2}a(5)^{2}$
$2u + 5a = 4$
For first $8\sec\ $of motion $s_{8} = 20\ metre$
$20 = 8u + \frac{1}{2}a(8)^{2} \Rightarrow 2u + 8a = 5$
By solving $u = \frac{7}{6}m/s$ 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 $2\ sec = s_{10} - s_{8}$
$= 28.3 - 20 = 8.3m$