A particle is moving in a straight line and passes through a point O with a velocity of 6m{s^{

English

Gujarati

Hindi

2.Motion in Straight Line

hard

A particle is moving in a straight line and passes through a point $O$ with a velocity of $6\;m{s^{ - 1}}.$ The particle moves with a constant retardation of $2\;m{s^{ - 2}}\;$ for $4$ s and there after moves with constant velocity. How long after leaving $O$ does the particle return to $O$..........$s$

$3$

$8$

Never

$4$

Solution

(b) Let the particle moves toward right with velocity $6\, m/s$. Due to retardation after time ${t_1}$ its velocity becomes zero.

From $v = u – a\,t$==> $0 = 6 – 2 \times {t_1}$ ==> ${t_1} = 3\,\sec $

But retardation works on it for $4 \,sec$. It means after reaching point $A$ direction of motion get reversed and acceleration works on the particle for next one second.

${S_{OA}} = u\,{t_1} – \frac{1}{2}a\,t_1^2$$ = 6 \times 3 – \frac{1}{2}(2)\;{(3)^2} = 18 – 9 = 9m$

${S_{AB}} = \frac{1}{2} \times 2 \times {(1)^2} = 1m$

$\therefore $ ${S_{BC}} = {S_{0A}} – {S_{AB}}$ $ = 9 – 1$$ = 8m$

Now velocity of the particle at point $B$ in return journey

$v = 0 + 2 \times 1$$ = 2m/s$

In return journey from $B$ to $C$, particle moves with constant velocity $2 m/s$ to cover the distance $8m$.

Time taken $ = \frac{{{\rm{Distance }}}}{{{\rm{Velocity}}}} = \frac{8}{2} = 4\sec $

Total time taken by particle to return at point 0 is ==>$T = {t_{0A}} + {t_{AB}} + {t_{BC}}$$ = 3 + 1 + 4 = 8\sec $.

Standard 11

Physics

A three-wheeler starts from rest, accelerates uniformly with $1\; m/s^{2}$ on a straight road for $10\; s$, and then moves with uniform velocity. Plot the distance covered by the vehicle during the $n ^{\text {th }}$ second $( n =1,2,3 \ldots .)$ versus $n$. What do you expect this plot to be during accelerated motion : a straight line or a parabola?

medium

A particle starts from rest at $x=0\; m$ with an acceleration of $1 \,m / s ^2$. At $t = 5\;s _{ s }$ it receives an additional acceleration in the same direction as its motion. At $t =10\; s$ its speed and position are $v$ and $x$, respectively. Had the additional acceleration not been provided, its speed and position would have been $v _0$ and $x _0$, respectively. It is found that $x – x _0$ is $12.5 \,m$. Then one can conclude that $v – v _0$ is ………….. $\,m / s$

normal

(KVPY-2021)

A particle starting from rest and moving with a uniform acceleration along a straight line covers distances $a$ and $b$ in successive intervals of $p$ and $q$ second. The acceleration of the particle is

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

hard

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

Solution

Similar Questions

A particle starting from rest and moving with a uniform acceleration along a straight line covers distances $a$ and $b$ in successive intervals of $p$ and $q$ second. The acceleration of the particle is

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 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).

Start a Free Trial Now