A boggy of uniformly moving train is suddenly | vedclass

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(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 tra

Vedclass · Accepted Answer

First will be half of second

Vedclass · Answer

(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 $\frac{{{S_b}}}{{{S_t}}} = \frac{1}{2}$

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

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