I seem to get a fair number of hits querying the basic formulas for velocity, so I’ve decided to post them.
Let [tex]\cal{O}[/tex] be an object moving along a path. The position of the object [tex]\cal{O}[/tex] is a function of the time [tex]t[/tex]. We denote the position vector [tex]\bold{x}(t)[/tex].
The time derivative of the position vector [tex]\bold{x}(t)[/tex] is the velocity vector [tex]\bold{v}(t)[/tex] of the object [tex]\cal{O}[/tex].
[tex]\bold{v}(t) \equiv \frac{d\bold{x}(t)}{dt} = \bold{\dot{x}}(t)[/tex].
The time derivative of the velocity vector [tex]\bold{v}(t)[/tex] is called the acceleration vector [tex]\bold{a}(t)[/tex] of the object [tex]\cal{O}[/tex] .
[tex]\bold{a}(t) \equiv \frac{d\bold{v}(t)}{dt} = \frac{d^2\bold{x}(t)}{dt^2} = \bold{\ddot{x}}(t) [/tex].
Integrate this last equation with respect to [tex]t[/tex],
[tex]\int \bold{\ddot{x}}(t) \ dt=\int \bold{a}(t) \ dt \implies \bold{\dot{x}}(t) =\bold{a}( t)t + \bold{C}[/tex],
where [tex]\bold{C}[/tex] is a constant vector to be determined by the initial conditions. Now, [tex]\bold{\dot{x}}(t) =\bold{v}(t)[/tex] so that
[tex]\bold{v}(t) =\bold{a}(t) t + \bold{C}[/tex]
We assume that at time [tex]t =0[/tex] the object [tex]\cal{O}[/tex] is moving with initial velocity [tex]\bold{v}(0) =\bold{v_0}[/tex]
[tex] \bold{v}(0) = \bold{v_0} =\bold{a}(0) \ 0 + C \implies C = \bold{v_0}[/tex]
and substitution gives the familiar formula for the velocity of an object
[tex]\bold{v}(t) = \bold{v_0} + \bold{a}(t) t[/tex].
We can integrate the equation
[tex]\bold{\dot{x}}(t) =\bold{v_0} + \bold{a}(t) t[/tex] with respect to [tex]t[/tex] a second time
[tex]\int \bold{\dot{x}}(t) \ dt = \int \bold{v_0} + \bold{a}(t) t \ dt[/tex] to give
[tex] \bold{x}(t) = \bold{v_0} t+ \frac{1}{2}\bold{a}(t) t^2 + \bold{K} [/tex], where [tex]\bold{K}[/tex] is a constant vector to be determined from the initial conditions.
We assume that at [tex]t =0[/tex] the object [tex]\cal{O}[/tex] has position vector [tex]\bold{x}(0)=\bold{x_0}[/tex]
[tex] \bold{x}(0) = \bold{x_0} = \bold{v_0} t+ \frac{1}{2}\bold{a}(0) 0^2 + \bold{K} \implies \bold{K}=\bold{x_0}[/tex]
and we have the familiar formula for the displacement (position) of the object at time [tex]t[/tex]
[tex] \bold{x}(t) = \bold{x_0} + \bold{v_0} t+ \frac{1}{2}\bold{a}(t) t^2 [/tex].
In a first treatment we often simplify things a bit. Firstly, to simplify notation we don’t write in the explicit time dependence for the position [tex]\bold{x}(t)[/tex] , so that it becomes [tex]\bold{x}[/tex] and similarly for [tex]\bold{v}[/tex] and [tex]\bold{a}[/tex]. Secondly we assume that the object starts from the origin rather than from some arbitrary position, thus we take [tex]\bold{x_0}=0[/tex], (which is really just a change of co-ordinates). This gives the familiar form of the equations of motion from high school text books:
[tex] \bold{x} = \bold{v_0} t+ \frac{1}{2}\bold{a} t^2 [/tex]
[tex]\bold{v} = \bold{v_0} + \bold{a} t[/tex].
These equations are commonly written as
[tex] s = u t+ \frac{1}{2} a t^2 [/tex]
[tex] v = u+ a t[/tex],
where we have used the alternate notation [tex]s \equiv x[/tex] for the position and [tex]u \equiv v_0[/tex] for the intial velocity. Note that we drop the bold face symbols [tex]\bold{v} \rightarrow v[/tex] which are used to denote vectors, since in a first treatment we consider only the one dimensional or scalar case.
Squaring the last equation gives
[tex]v^2 = (u+ a t)^2 = u^2 + 2 u a t + a^2t^2 = u^2 + 2a (ut+\frac{1}{2}at^2) [/tex]
[tex]\phantom{v^2}= u^2 + 2as[/tex]
another commonly used equation in high school physics texts.
The average velocity over the time interval [tex] (t_1,t_2) \equiv (t_i,t_f)[/tex] is given by
[tex]v_{av} = \frac{s_2 -s_1}{t_2-t_1} = \frac{s_f-s_i}{t_f-t_i} = \frac{\Delta(s)}{\Delta(t)}[/tex]
where, [tex] s_1 \equiv s_i[/tex] is the intial displacement and [tex]s_2 \equiv s_f[/tex] is the final displacement. Finally, in words: it is the change in position [tex]\Delta(s)[/tex] divided by the change in time [tex]\Delta(t)[/tex].