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 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
Both will be equal
First will be half of second
First will be $1/4$ of second
No definite ratio
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
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 $\frac{{{S_b}}}{{{S_t}}} = \frac{1}{2}$
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