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(B) Let the particle move toward the right with velocity $u = 6\;m/s$. Due to retardation $a = 2\;m/s^2$,its velocity becomes zero after time $t_1$.Fr

Vedclass · Accepted Answer

$8\;s$

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(B) Let the particle move toward the right with velocity $u = 6\;m/s$. Due to retardation $a = 2\;m/s^2$,its velocity becomes zero after time $t_1$.From $v = u - at$,we have $0 = 6 - 2t_1$,which gives $t_1 = 3\;s$.Since the retardation acts for $4\;s$,the particle continues to accelerate in the opposite direction for the remaining $1\;s$ $(4\;s - 3\;s = 1\;s)$.Distance covered in the first $3\;s$ (from $O$ to $A$): $S_{OA} = ut_1 - \frac{1}{2}at_1^2 = 6(3) - \frac{1}{2}(2)(3)^2 = 18 - 9 = 9\;m$.Distance covered in the next $1\;s$ (from $A$ to $B$): $S_{AB} = \frac{1}{2}at^2 = \frac{1}{2}(2)(1)^2 = 1\;m$.Position of $B$ relative to $O$: $S_{OB} = S_{OA} - S_{AB} = 9 - 1 = 8\;m$.Velocity at $B$ (in return journey): $v_B = at = 2(1) = 2\;m/s$.From $B$ onwards,the particle moves with a constant velocity of $2\;m/s$ to cover the distance $S_{OB} = 8\;m$.Time taken to travel from $B$ to $O$: $t_{BO} = \frac{S_{OB}}{v_B} = \frac{8}{2} = 4\;s$.Total time taken to return to $O$: $T = t_{OA} + t_{AB} + t_{BO} = 3 + 1 + 4 = 8\;s$.

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

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