$\vec{u}$ may be perpendicular or parallel to $\vec{v}$ depending on whether it is a transverse or longitudinal wave.

$y=A \sin k x-\omega t$

$\frac{d y}{d t}=A \omega \cos k x-\omega t$

$\frac{d y}{d t}=v$

$u=-A \omega \cos k x-\omega t$

$|v|=\frac{\omega}{k}$

Slope of $y=A \sin k x-\omega t$

$\frac{d y}{d x}=A k \cos k x-\omega t$

$\left|\frac{d y}{d x}\right|=|v|\left|\frac{d y}{d x}\right|$

$u=|v| \text { (Slope of waveform) }$