The velocity of a particle, denoted $\mathbf{v}$ is a vector in the direction of motion. Its magnitude is therefore the speed of the particle.
If a particle is moving for $t$ seconds with constant velocity $\mathbf{v}\mathrm{ms^{-1} }$ it will have moved $\mathbf{v}t\mathrm{m}$. The displacement is parallel to the velocity. The magnitude of the displacement is the distance from the starting point.
What is the speed of a particle that is moving with constant velocity $\mathbf{v} = (4\mathbf{i} + 2\mathbf{j}) ms^{-1}$?
The speed is the magnitude of the velocity so we have \begin{align} \mathbf{v} & = (4\mathbf{i} + 2\mathbf{j}), \\ \lvert \mathbf{v} \rvert & = \sqrt{ 4^2 + 2^2}, \\ & = 2 \sqrt{5}, \\ & = 4.47 \mathrm{ms^{-1} } \text{ (to 3s.f.).} \end{align}
What is the distance moved every $8 \mathrm{seconds}$ by a particle that is moving with constant velocity $\mathbf{v} = (5\mathbf{i} - \mathbf{j})$?
Here we can use that $\text{displacement} = \mathbf{v} \times t$ to find the distance moved. \begin{align} \text{Displacement} & = \mathbf{v}t, \\ & = (5 \mathbf{i} - \mathbf{j}) \times 8, \\ & = (40 \mathbf{i} - 8 \mathbf{j})\mathrm{m}, \\ \text{Distance} & = \lvert \text{Displacement} \rvert, \\ & = \lvert \mathbf{v} t \rvert, \\ & = \sqrt{40^2 + 8^2}, \\ & = 8 \sqrt{26}, \\ & = 40.792\mathrm{m}. \end{align} Another method would be to calculate the speed and multiply this by the time. \begin{align} \text{speed} & = \sqrt{5^2 + 1^2}, \\ & = \sqrt{26}\mathrm{ms^{-1} }, \\ \text{distance} & = \text{speed} \times \text{time}, \\ & = \sqrt{26} \times 8, \\ & = 40.792\mathrm{m}. \end{align}
If a particle starts from the point with positive vector $\mathbf{r_0}$ and moves with constant velocity $\mathbf{v}$, then it's initial position at time $t$ is $\mathbf{v}t$ and it's position vector $\mathbf{r} = \mathbf{r_0} + \mathbf{v}t$.
Suppose that a particle is at a point with position vector $(3\mathbf{i} - 7\mathbf{j})m$ at time $t=0$ and is moving at a constant velocity. Five seconds later it is at a point with position vector $(6 \mathbf{i} + 7\mathbf{j})m$. What is the velocity of the particle?
We use the formula $\mathbf{r} = \mathbf{r_0} + \mathbf{v}t$ to find the velocity $\mathbf{v}$. We begin by finding the displacement.
\begin{align} \text{Displacement} & = (6 \mathbf{i} + 7 \mathbf{j} ) - (3 \mathbf{i} - 7 \mathbf{j}), \\ & = (6 - 3) \mathbf{i} + (7 - (-7))\mathbf{j}, \\ & = \left(3 \mathbf{i} + 14 \mathbf{j}\right)\mathrm{m}, \end{align}
Therefore the particle travels $3 \mathbf{i} + 14 \mathbf{j}$ in $5$ seconds, and as it is travelling at a constant velocity we divide the displacement by the time taken in order to obtain the velocity $\mathbf{v}$.
\begin{align} \mathbf{v} & = \frac{1}{5} \left( 3 \mathbf{i} + 14 \mathbf{j} \right), \\ & = ( 0.6 \mathbf{i} + 2.8 \mathbf{j} )\mathrm{ms^{-1} }. \end{align}