(b) Let $'a'$ be the retardation of boggy then distance covered by it be $S$. If $u$ is the initial velocity of boggy after detaching from train (i.e. uniform speed of train)

${v^2} = {u^2} + 2as \Rightarrow 0 = {u^2} – 2as \Rightarrow {s_b} = \frac{{{u^2}}}{{2a}}$

Time taken by boggy to stop

$v = u + at \Rightarrow 0 = u – at \Rightarrow t = \frac{u}{a}$

In this time t distance travelled by train$ = {s_t} = ut = \frac{{{u^2}}}{a}$

Hence ratio $\frac{{{S_b}}}{{{S_t}}} = \frac{1}{2}$