A boggy of uniformly moving train is suddenly detached from train and stops after covering some dist

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2.Motion in Straight Line

hard

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

Solution

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

Standard 11

Physics

A train moves from rest with a uniform acceleration $a$ . Attaining a maximum speed $v$ it starts moving with uniform retardation $a$ . Assuming $s$ = total distance covered in the unidirectional motion of the train, its total time of journey and maximum speed are (respectively).

medium

$x-t$ graph for a uniformly accelerated particle is as shown in the figure. Then find the average velocity between points $(i)$ and $(ii)$ ……… $ms^{-1}$

hard

The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) \mathrm{s}$ are $125 \mathrm{~m}$ and $50 \mathrm{~m} / \mathrm{s}$, respectively. The distance travelled by the particle in $(\mathrm{t}+2)^{\mathrm{th}} \mathrm{s}$ is__________ $\mathrm{m}$.

hard

(JEE MAIN-2024)

Draw and explain the $v \to t$ graphs for uniformly accelerated motion.

medium

A particle returns to the starting point after $10\,s$. If the rate of change of velocity during the motion is constant, then its location after $7\,s$ will be same as that after $………..\,s$

medium

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

Solution

Similar Questions

A train moves from rest with a uniform acceleration $a$ . Attaining a maximum speed $v$ it starts moving with uniform retardation $a$ . Assuming $s$ = total distance covered in the unidirectional motion of the train, its total time of journey and maximum speed are (respectively).

$x-t$ graph for a uniformly accelerated particle is as shown in the figure. Then find the average velocity between points $(i)$ and $(ii)$ ……… $ms^{-1}$

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A particle returns to the starting point after $10\,s$. If the rate of change of velocity during the motion is constant, then its location after $7\,s$ will be same as that after $………..\,s$

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