3. At time \(t = 0\), a particle of mass \(m\) is projected vertically downwards with speed \(U\) from a point above the ground. At time \(t\) the speed of the particle is \(v\) and the magnitude of the air resistance is modelled as being \(m k v\), where \(k\) is a constant. Given that \(U < \frac { \boldsymbol { g } } { \mathbf { 2 } \boldsymbol { k } }\), find, in terms of \(k , U\) and \(g\), the time taken for the particle to double its speed.
(8)

## Question 3:

| Working/Answer | Marks | Notes |
|---|---|---|
| $mg - mkv = m\frac{\mathrm{d}v}{\mathrm{d}t}$ | M1* A1 A1 | |
| $\int \mathrm{d}t = \int \frac{\mathrm{d}v}{g-kv}$ | DM1* | |
| $t = -\frac{1}{k}\ln(g-kv)+c$ | A1cao | |
| $t=0, v=u \Rightarrow c = \frac{1}{k}\ln(g-ku)$ | M1† | |
| $T = \frac{1}{k}\ln(g-ku) - \frac{1}{k}\ln(g-2ku)$ | DM1† | |
| $= \frac{1}{k}\ln\!\left(\dfrac{g-ku}{g-2ku}\right)$ | A1 | **Total: 8** |

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3. At time $t = 0$, a particle of mass $m$ is projected vertically downwards with speed $U$ from a point above the ground. At time $t$ the speed of the particle is $v$ and the magnitude of the air resistance is modelled as being $m k v$, where $k$ is a constant.

Given that $U < \frac { \boldsymbol { g } } { \mathbf { 2 } \boldsymbol { k } }$, find, in terms of $k , U$ and $g$, the time taken for the particle to double its speed.\\
(8)\\

\hfill \mbox{\textit{Edexcel M4 2008 Q3 [8]}}