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A particle moves according to the law $s=t^{3}-6 t^{2}+9 t+25 .$ The displacement of
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27 units
Given $s=t^{3}-6 t^{2}+9 t+25$ ....(1)
$$
\begin{array}{l}v=\frac{d s}{d t}=3 t^{2}-12 t+9 \\ \frac{d v}{d t}=6 t-12\end{array}
$$
Given $6 t-12=0 \Rightarrow t=2$
$\therefore$ from $(1), s=(2)^{3}-6 \times 4+18+25=27$
$$
\begin{array}{l}v=\frac{d s}{d t}=3 t^{2}-12 t+9 \\ \frac{d v}{d t}=6 t-12\end{array}
$$
Given $6 t-12=0 \Rightarrow t=2$
$\therefore$ from $(1), s=(2)^{3}-6 \times 4+18+25=27$
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