Explanation

(b) Let $s_{1}$ be the distance travelled by train with acceleration $1{ms}^{- 2}$ for time $t_{1}and\ s_{2}$ be the distance travelled by train with retardation $3\ {ms}^{- 2}$ for time $t_{2}.$ If $v$ is the velocity of train after time $t_{1},$ then
$\ \ \ \ \ \ \ \ \ \ v = 1 \times t_{1}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
And $s_{1} = \frac{1}{2} \times 1 \times t_{1}^{2} = \frac{t_{1}^{2}}{2}\ \ \ \ \ \ \ \ \ \ldots(ii)$
$\ \ \ \ \ \ \ \ \ \ \ v = 3t_{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
And $s_{2} = vt_{2} - \frac{1}{2} \times 3 \times t_{2}^{2}$
From Eqs. (i) and (iii),
$\ \ \ \ \ \ \ \ \ \ \ \ t_{1} = {3t}_{2}or\ t_{2} = \frac{t_{1}}{3}$
$\therefore s_{1} + s_{2} = \frac{t_{1}^{2}}{2} + t_{1} \times \frac{t_{1}}{3} - \frac{3}{2} \times \frac{t_{1}^{2}}{9}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{2}{3}t_{1}^{2}$
$\ \ \ \ \ \ \ \ \ 1215 = \frac{2}{3}t_{1}^{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{1} = \sqrt{\frac{3 \times 1215}{2}} = 42.69\ s$
Total time $= t_{1} + t_{2} = t_{1} + \frac{t_{1}}{3} = 56.9\ s$