Question

A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The ratio of distance covered by the boggy and distance covered by the train when the boggy has stopped is

• A. $1:1$
• B. $1:2$
• C. $1:4$
• D. $3:2$

Answer 1

The correct option is B $1:2$


$a=$ retardation for the boggy

Using $v=u+at$,
$0=u-at$, taking $-ve$ sign for retardation.
$t=u/a......\left(1\right)$

Let $S_t$ and $S_b$ be the distances coverd by train and boggy respectively.

$S_t=ut=u\times \frac{u}{a}=u^2/a$

Using, $v^2=u^2-2a{S}_b$
$0=u^2-2a{S}_b$
$S_b=\frac{u^2}{2a}$
$\Rightarrow S_b=S_t/2$
$\Rightarrow \frac{S_b}{S_t}=\frac{1}{2}$