**Question:**

A particle starts with speed $\displaystyle 20 \ ms^{-1}$ and moves in a straight line. The particle is subject to a retardation which is initially $\displaystyle 5 \ ms^{-2}$ and which increases uniformly with the distance moved, having a value of $\displaystyle 11 \ ms^{-2}$ when the particle has moved $\displaystyle 12 \ m$. Given that the particle has speed $\displaystyle v \ ms^{-1}$ and has moved a distance of $\displaystyle x$, show that, while the particle is in motion

$\displaystyle v \frac{\mathrm{d}v}{\mathrm{d}x} = -\left(5+ \frac12 x\right)$.

**My Problem with this question:**
I can't derive it.

Help please? Thanks in advance.