A particle \(P\) moves along the \(x\)-axis. At time \(t\) seconds the velocity of \(P\) is \(v \text{ ms}^{-1}\) in the positive \(x\)-direction, where \(v = 3t^2 - 4t + 3\). When \(t = 0\), \(P\) is at the origin \(O\). Find the distance of \(P\) from \(O\) when \(P\) is moving with minimum velocity. [8]

$\frac{dv}{dr} = 6t - 4$ | M1 A1 |
$6t - 4 = 0 \Rightarrow t = \frac{2}{3}$ | M1 A1 |
$s = \int 3t^2 - 4t + 3 \, dt = t^3 - 2t^2 + 3t (+c)$ | M1 A1 |
$t = \frac{2}{3} \Rightarrow s = -\frac{16}{27} + 2$ so distance is $\frac{38}{27}$ m | M1 A1 |

**Total: [8]**

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A particle $P$ moves along the $x$-axis. At time $t$ seconds the velocity of $P$ is $v \text{ ms}^{-1}$ in the positive $x$-direction, where $v = 3t^2 - 4t + 3$. When $t = 0$, $P$ is at the origin $O$. Find the distance of $P$ from $O$ when $P$ is moving with minimum velocity.
[8]

\hfill \mbox{\textit{Edexcel M2 2010 Q1 [8]}}