Question

A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation

• A. Both will be equal
• B. Boggy will cover half of the distance covered by train
• C. Boggy will cover quarter of the distance covered by train
• D. No definite ratio

Answer 1

The correct option is B Boggy will cover half of the distance covered by train

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), then we have
$v^2=u^2+2as\Rightarrow 0=u^2-2a{s}_b\Rightarrow s_b=\frac{u^2}{2a}$
Time take by boggy to stop
$\Rightarrow v=u+at\Rightarrow 0=u-at\Rightarrow t=\frac{u}{a}$
In this time $t$ distance travelled by train is
$s_t=ut=\frac{u^2}{a}$
Hence, the ratio is
$\frac{s_b}{s_t}=\frac{\left(\frac{u^2}{2a}\right)}{\left(\frac{u^2}{a}\right)}=\frac{1}{2}\Rightarrow s_b=0.5s_t$
Thus, the distance covered by the boggy will be half of the distance covered by train.