A bogie of a uniformly moving train is sudden | vedclass

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(B) Let $a$ be the retardation of the bogie,and $S_b$ be the distance covered by it before it stops. Let $u$ be the initial velocity of the bogie at t

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The first will be half of the second

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(B) Let $a$ be the retardation of the bogie,and $S_b$ be the distance covered by it before it stops. Let $u$ be the initial velocity of the bogie at the moment it detaches (which is equal to the uniform speed of the train).Using the equation of motion $v^2 = u^2 + 2as$,where final velocity $v = 0$:$0 = u^2 - 2aS_b \Rightarrow S_b = \frac{u^2}{2a}$Now,calculate the time $t$ taken by the bogie to stop using $v = u + at$:$0 = u - at \Rightarrow t = \frac{u}{a}$In this same time $t$,the train continues to move with uniform velocity $u$. Therefore,the distance covered by the train is:$S_t = u 	imes t = u 	imes \frac{u}{a} = \frac{u^2}{a}$Comparing the two distances:$\frac{S_b}{S_t} = \frac{u^2 / 2a}{u^2 / a} = \frac{1}{2}$Thus,the distance covered by the bogie is half the distance covered by the train.

Column $I$	Column $II$
$(A)$ Constant positive acceleration	$(p)$ Speed may increase
$(B)$ Constant negative acceleration	$(q)$ Speed may decrease
$(C)$ Constant displacement	$(r)$ Speed is zero
$(D)$ Constant slope of $a-t$ graph	$(s)$ Speed must increase

$A$ bogie of a uniformly moving train is suddenly detached from the train and stops after covering some distance. What is the relation between the distance covered by the bogie and the distance covered by the train in the same time?

Similar Questions

Match the following columns.

Column $I$ Column $II$

$(A)$ Constant positive acceleration $(p)$ Speed may increase

$(B)$ Constant negative acceleration $(q)$ Speed may decrease

$(C)$ Constant displacement $(r)$ Speed is zero

$(D)$ Constant slope of $a-t$ graph $(s)$ Speed must increase

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