b
(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 \(\frac{{{S_b}}}{{{S_t}}} = \frac{1}{2}\)