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(C) Given the displacement function $S(t) = t^3 - 16t^2 + 64t - 16$. To find the maximum displacement,we first find the velocity $v(t)$ by differentia

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$\frac{8}{3}$

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(C) Given the displacement function $S(t) = t^3 - 16t^2 + 64t - 16$. To find the maximum displacement,we first find the velocity $v(t)$ by differentiating $S$ with respect to $t$: $v(t) = \frac{dS}{dt} = 3t^2 - 32t + 64$. Setting $v(t) = 0$ to find the critical points: $3t^2 - 32t + 64 = 0$. Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{32 \pm \sqrt{(-32)^2 - 4(3)(64)}}{2(3)} = \frac{32 \pm \sqrt{1024 - 768}}{6} = \frac{32 \pm \sqrt{256}}{6} = \frac{32 \pm 16}{6}$. This gives two values: $t_1 = \frac{48}{6} = 8$ and $t_2 = \frac{16}{6} = \frac{8}{3}$. Now,we find the second derivative to check for maxima: $a(t) = \frac{d^2S}{dt^2} = 6t - 32$. For $t = 8$: $a(8) = 6(8) - 32 = 48 - 32 = 16 > 0$ (Local minimum). For $t = \frac{8}{3}$: $a(\frac{8}{3}) = 6(\frac{8}{3}) - 32 = 16 - 32 = -16 Thus,the displacement is maximum at $t = \frac{8}{3}$.

The displacement $S$ of a particle measured from a fixed point $O$ on a line is given by $S = t^3 - 16t^2 + 64t - 16$. Then the time at which displacement of the particle is maximum is

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