2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt { \frac { g } { k } }\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(m k v ^ { 2 }\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed.
(9)

**Part (a)**
$-mg - mkv^2 = ma$

$-(g + kv^2) = v\frac{dv}{dx}$

| M1 A1 M1 |

**Part (b)**
$\pm \int dx = \int \frac{-vdv}{(g+kv^2)}$

$X = \frac{1}{2k}\left[\ln(g+kv^2)\right]_{\frac{x}{\sqrt{k}}}^{\frac{\sqrt{k}}{k}}$

$= \frac{1}{2k}\left(\ln 2g - \ln \frac{5g}{4}\right)$

$= \frac{1}{2k}\ln \frac{8}{5}$

| DM1 A1 (both previous) M1 A1 M1 A1 |

**Total: [9]**

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2. At time $t = 0$, a particle $P$ of mass $m$ is projected vertically upwards with speed $\sqrt { \frac { g } { k } }$, where $k$ is a constant. At time $t$ the speed of $P$ is $v$. The particle $P$ moves against air resistance whose magnitude is modelled as being $m k v ^ { 2 }$ when the speed of $P$ is $v$. Find, in terms of $k$, the distance travelled by $P$ until its speed first becomes half of its initial speed.\\
(9)

\hfill \mbox{\textit{Edexcel M4 2009 Q2 [9]}}