Explanation

(d) The braking retardation will remain same and assumed to be constant, let it be $a.$
From the 3$^{rd}$ equation of motion, $v^{2} = u^{2} + 2as$
Ist case :$\ \ \ \ \ \ \ \ \ 0 = \left( 60 \times \frac{5}{18} \right)^{2} - 2a \times s_{1}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s_{1} = \frac{(60\ \times 5/18)^{2}}{2a}$
IInd case : $\ \ \ \ 0 = \left( 120 \times \frac{5}{18} \right)^{2} - 2a \times s_{2}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s_{2} = \frac{(120\ \times 5/18)^{2}}{2a}$
$\therefore\frac{s_{1}}{s_{2}} = \frac{1}{4}\ \Rightarrow s_{2} = 4s_{1} = 4 \times 20 = 80\ m$