A particle \(P\) moves on the positive \(x\)-axis. When the distance of \(P\) from the origin \(O\) is \(x\) metres, the acceleration of \(P\) is \((7 - 2x)\) m s\(^{-2}\), measured in the positive \(x\)-direction. When \(t = 0\), \(P\) is at \(O\) and is moving in the positive \(x\)-direction with speed 6 m s\(^{-1}\). Find the distance of \(P\) from \(O\) when \(P\) first comes to instantaneous rest. [6]

$v = \frac{dv}{dx} = 7 - 2x$ | M1 |
$\frac{1}{2}v^2 = 7x - x^2 (+c)$ | M1A1 |
$x = 0, v = 6 \Rightarrow c = 18$ | A1 |
$v = 0: x^2 - 7x - 18 = 0$ | M1 |
$(x+2)(x-9) = 0$ | |
$\therefore x = 9$ | A1 |

A particle $P$ moves on the positive $x$-axis. When the distance of $P$ from the origin $O$ is $x$ metres, the acceleration of $P$ is $(7 - 2x)$ m s$^{-2}$, measured in the positive $x$-direction. When $t = 0$, $P$ is at $O$ and is moving in the positive $x$-direction with speed 6 m s$^{-1}$. Find the distance of $P$ from $O$ when $P$ first comes to instantaneous rest.
[6]

\hfill \mbox{\textit{Edexcel M3 2011 Q1 [6]}}