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Question

(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\,

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(b) Let the particle moves toward right with velocity $6\, m/s$. Due to retardation after time ${t_1}$ its velocity becomes zero.

From $v = u - a\,t$==> $0 = 6 - 2 	imes {t_1}$ ==> ${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 	imes 3 - \frac{1}{2}(2)\;{(3)^2} = 18 - 9 = 9m$

${S_{AB}} = \frac{1}{2} 	imes 2 	imes {(1)^2} = 1m$

$	herefore $ ${S_{BC}} = {S_{0A}} - {S_{AB}}$ $ = 9 - 1$$ = 8m$

Now velocity of the particle at point $B$ in return journey

$v = 0 + 2 	imes 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{{{m{Distance }}}}{{{m{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 $.

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