How is the magnitude of a direction vector equal to speed? In thomas calculus $12$ edition it gives the position vector along a line as
$\vec r(t) = \vec r_0  + t\vec v$
where $\vec r_0$ is the initial position vector and $\vec v$ is the direction vector.
Then it rewrites it as 
$\vec r(t) = \vec r_0 + t, |\vec v| \, \frac{\vec v}{|\vec v|}$
and it refers to $|\vec v|$ as the speed, which I don't understand. $|\vec v|$ is basically the length of the direction vector. Wouldn't speed be the $y$ component of the direction vector divided by the $x$ component, that is the slope of the position $\vec v$ time graph formed by $\vec r(t)$.

The position vector is described parametrically with parameter $t$ as 
$$\vec r(t)=\vec r_0+\vec v\,t$$
Note that if we wish to measure the rate of change of position with respect to the parameter $t$, we have
$$\frac{d\vec r(t)}{dt}=\vec v$$
The magnitude of that rate of change is 
$$\left|\frac{d\vec r(t)}{dt}\right|=|\vec v|$$
Therefore, we can write
$$\begin{align}
\vec r(t)&=\vec r_0+\left(|\vec v|\,t\right)\,\left(\frac{\vec v|}{|\vec v|}\right)\\\\
&=\vec r_0 +t\,\left|\frac{d\vec r(t)}{dt}\right|\,\left(\frac{\vec v}{|\vec v|}\right)\tag 1
\end{align}$$
Note that in $(1)$, the term $\frac{\vec v}{|\vec v|}$ is a unit vector that points in the direction of $\vec v=\frac{d\vec r(t)}{dt}$ and $\left|\frac{d\vec r(t)}{dt}\right|$ is the magnitude of $\vec v$, which is the magnitude of the rate of change of the position vector with respect to the parameter $t$.
Now, interpreting $t$ as time and $\frac{d\vec r(t)}{dt}$ as velocity, we see that the position vector is given in terms of its initial position, $\vec r_0$, plus a vector that points in the direction of the velocity with magnitude equal to the product of the speed, $|\vec v|$, and time, $t$.
