The relation 3t = sqrt{3x} + 6 describes the | vedclass

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(D) Given the relation: $3t = \sqrt{3x} + 6$.Rearranging for $\sqrt{3x}$: $\sqrt{3x} = 3t - 6$.Squaring both sides: $3x = (3t - 6)^2 = 9t^2 - 36t + 36

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(D) Given the relation: $3t = \sqrt{3x} + 6$.Rearranging for $\sqrt{3x}$: $\sqrt{3x} = 3t - 6$.Squaring both sides: $3x = (3t - 6)^2 = 9t^2 - 36t + 36$.Dividing by $3$: $x = 3t^2 - 12t + 12$.Velocity $v$ is the derivative of displacement $x$ with respect to time $t$: $v = \frac{dx}{dt} = \frac{d}{dt}(3t^2 - 12t + 12) = 6t - 12$.For velocity to be zero: $6t - 12 = 0 \Rightarrow t = 2 \ s$.Substituting $t = 2 \ s$ into the displacement equation: $x = 3(2)^2 - 12(2) + 12 = 12 - 24 + 12 = 0 \ m$.

The relation $3t = \sqrt{3x} + 6$ describes the displacement of a particle in one direction,where $x$ is in $m$ and $t$ is in $s$. The displacement,when velocity is zero,is ......... $m$.

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