Explanation

(a) Le the initial velocity $= u$
And acceleration $= a$
In Ist case$\mathbf{\ \ \ \ \ \ \ \ \ }s_{1} = ut_{1} + \frac{1}{2}\ at_{1}^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 200 = 2u + 2a\ \ \ \ \ \ \ \ \ \ \ \left( \because t_{1} = 2\ s \right)$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ u + a = 100\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
In IInd case
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s_{2} = ut_{2} + \frac{1}{2}\ at_{2}^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 420 = 6u + 18a\ \ \ \ \ \ \left( \because t_{2} = 2 + 4 = 6\ s \right)$
Or $\ \ \ \ \ \ \ \ \ \ \ \ 3a + u = 70\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Solving Eqs. (i) and (ii), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a = - 15\ {ms}^{- 2}$
And $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u = 115\ {ms}^{- 1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = u + at$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 115 - 15 \times 7 = 10\ {ms}^{- 1}$