Explanation

(c) We know that,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s_{t} = u + \frac{1}{2}a(2t - 1)$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 65 = u + \frac{1}{2}a\lbrack 10 - 1\rbrack$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 65 = u + \frac{9}{2}\ a\ \ \ \ \ \ \ \ \ \ \ldots(i)$
Again, $t = 9s\ \ and\ \ s = 105\ m$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 105 = u + \frac{1}{2}a\lbrack 18 - 1\rbrack$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ 105 = u + \frac{17}{2}a\ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
On substracting Eq. (i) from Eq. (ii)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ 105 - 65 = \frac{17}{2}a - \frac{9}{2}a$
Or $40 = \frac{8}{2}\ a$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 40 = 4a$
Or $a = 10\ {ms}^{- 2}$
On putting the value of $a$ in Eq. (i),
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 65 = u + \frac{9}{2} \times 10$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 65 = u + 45$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u = 20\ {ms}^{- 1}$
Now we know that from second equation of motion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = ut + \frac{1}{2}\ at^{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ s = 20 \times 20 + \frac{1}{2} \times 10 \times (20)^{2}$
Or $\ \ \ \ \ \ \ \ \ \ \ s = 400 + 5 \times 400$
Or $\ \ \ \ \ \ \ \ \ \ \ s = 2400\ m$