A particle of mass \(m\) kg moves in a horizontal straight line. Its initial speed is \(u\) ms\(^{-1}\) and the only force acting on it is a variable resistance of magnitude \(mkv\) N, where \(v\) ms\(^{-1}\) is the speed of the particle after \(t\) seconds and \(k\) is a constant. Show that \(v = ue^{-kt}\). [7 marks]

$m\frac{dv}{dt} = -mky$ | M1 A1 M1 A1 | 
$\frac{1}{v}dv = \int -k dt$ | M1 A1 A1 | 
$\ln v = -kt + c$ | | 
$\ln u = c$, so $\ln \frac{v}{u} = -kt$ | | 
$v = ue^{-kt}$ | | 

**Total: 7 marks**

A particle of mass $m$ kg moves in a horizontal straight line. Its initial speed is $u$ ms$^{-1}$ and the only force acting on it is a variable resistance of magnitude $mkv$ N, where $v$ ms$^{-1}$ is the speed of the particle after $t$ seconds and $k$ is a constant.

Show that $v = ue^{-kt}$. [7 marks]

\hfill \mbox{\textit{Edexcel M3  Q1 [7]}}