Derivation of Kinematics Equation

Question

$\vec{s}$ is an arbitrary position vector, which can change with time.

$\vec{v} = \dfrac{\Delta \vec{s}}{\Delta t}$ is the velocity vector.

$\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t}$ is the acceleration vector, and it readily follows that

$\vec{v_f}=\vec{v_i}+\vec{a}\Delta t$

$\dfrac{\Delta \vec{s}}{\Delta t}=\vec{v_i}+\vec{a}\Delta t$

$\Delta \vec{s}=\vec{v_i}\Delta t+\vec{a}(\Delta t)^2$

However this differs from the accepted $\Delta x=v_0t+\frac{1}{2}at^2$

What's wrong with my reasoning and how can I get to the correct answer using purely vectors?

Answer 1

Your second to last step is the statement $$\frac{\Delta{\bf s}}{\Delta t}={\bf v}_f$$ which doesn't make a lot of sense. One could argue that it makes more sense to use the velocity at the middle of the time interval $\Delta t$, in which case you get the desired result.