Q. 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) First will be half of second
C) First will be 1/4 of second
D) No definite ratio
Explanation
(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 $\frac{s_{b}}{s_{1}} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.