A block is thrown with a velocity of 2 , m/s | vedclass

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(D) Let the belt move towards the left with $v_b = -4 \, m/s$ and the block be thrown towards the right with $v_{block,g} = 2 \, m/s$.Relative velocit

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(D) Let the belt move towards the left with $v_b = -4 \, m/s$ and the block be thrown towards the right with $v_{block,g} = 2 \, m/s$.Relative velocity of the block with respect to the belt is $v_{block,b} = v_{block,g} - v_b = 2 - (-4) = 6 \, m/s$.The block stops slipping on the belt after $t = 4 \, s$,meaning its final velocity relative to the belt is $v_f = 0$.Using $v_f = u + at$,we get $0 = 6 + a(4)$,so $a = -1.5 \, m/s^2$ (relative to the belt).Displacement relative to the belt: $S_b = u_{rel}t + 0.5at^2 = 6(4) + 0.5(-1.5)(4^2) = 24 - 12 = 12 \, m$.Displacement of the belt relative to the ground: $S_{belt,g} = v_b 	imes t = -4 	imes 4 = -16 \, m$.Displacement of the block relative to the ground: $S_{block,g} = S_b + S_{belt,g} = 12 - 16 = -4 \, m$. The magnitude is $4 \, m$.For displacement relative to the ground to be zero: $S_{block,g}(t) = (v_{block,g})t + 0.5a_g t^2 = 0$. Since $a_g = a = -1.5 \, m/s^2$,$2t - 0.75t^2 = 0 \Rightarrow t(2 - 0.75t) = 0 \Rightarrow t = 2/0.75 = 8/3 \, s$.

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