(c) $\because {S_1} = ut + \frac{1}{2}a{t^2}$…..(i)

and velocity after first $t \,sec$

$v = u + at$

Now, ${S_2} = vt + \frac{1}{2}a{t^2}$

$ = (u + at)t + \frac{1}{2}a{t^2}$….. (ii)

Equation (ii) -(i)

$\Rightarrow {S_2} – {S_1} = a{t^2}$

$ \Rightarrow a = \frac{{{S_2} – {S_1}}}{{{t^2}}} = \frac{{65 – 40}}{{{{(5)}^2}}} = 1\;m/{s^2}$

From equation (i), we get,

${S_1} = ut + \frac{1}{2}a{t^2}$ $ \Rightarrow 40 = 5u + \frac{1}{2} \times 1 \times 25$

$ \Rightarrow 5u = 27.5$

$\therefore \;\;u = 5.5\;m/s$