I was curious if the work-energy and impulse-momentum principles be combined where the common factor is average Force.
Work Energy: $F_{avg} \times d = 0.5m(v_f-v_0)^2$ Impulse Momentum: $F_{avg} \times t = m(v_f-v_0)$
Combining them you get $0.5tm(v_f-v_0)^2 = dm(v_f-v_0)$. Masses cancel each other leaving: $0.5t(v_f-v_0)^2 = d(v_f-v_0)$. This further reduces to $0.5t(v_f-v_0)=d$ or $(v_f-v_0)=\frac{2d}{t}$.
But wouldnt $v$ be equal to $\frac{d}{t}$ and not $v=\frac{2d}{t}$.
So yes, you can do this. It is seen to more fully be valid using techniques that we call "vectors" and "calculus", where one discovers that Newton's law $\sum_i \vec F_i = m \vec a = m \frac{d\vec v}{dt}$ actually implies, if you define the "power exerted by a force $\vec F_i$" as $P_i = \vec F_i\cdot\vec v$, that $$\sum_i P_i = \sum_i \vec F_i \cdot \vec v = \left(\sum_i ~\vec F_i\right) \cdot \vec v = m~\frac{d\vec v}{dt}\cdot \vec v = \frac12 m \frac{d}{dt}\left(\vec v\cdot\vec v\right) = \frac{dK}{dt},$$where $K = \frac12 m v^2$ and we're assuming that the mass is not changing over time.
That's probably a bit over your head at this stage, but the important thing is, yes, we can do this explicitly using only algebra for the special case of one constant force in the direction of motion.
In this case one has two times $t_0, t_1$ that one is interested in, two velocities $v_0, v_1$, and two positions, $x_0, x_1,$ and one acceleration. For constant accelerations, the average velocity can be expressed in two ways, $$v_\text{avg} = \frac{v_1 + v_0}{2} = \frac{x_1 - x_0}{t_1-t_0},$$but we also know that the acceleration is the change in velocity over change in time, $$a = \frac{v_1 - v_0}{t_1 - t_0}.$$ We can use these both to solve for and eliminate the time difference $t_1 - t_0$ as$$t_1 - t_0 = \frac{v_1 - v_0}{a},$$so our earlier expression says:$$\frac{v_1 + v_0}{2} = a~\frac{x_1 - x_0}{v_1 - v_0}.$$Multiplying through by $v_1 - v_0$ on both sides the left hand side is a classic expression for factoring the difference of squares: if you FOIL it out you will find that $(a + b)(a - b) = a^2 +ab - ab - b^2=a^2-b^2.$
Similarly here you will end up finding, $$\frac12 \big(v_1^2 - v_0^2\big) = a~(x_1 - x_0).$$Multiplying through by the mass and defining $K = \frac12 m~v^2$ one finds that $F=ma$ becomes $$K_1 - K_0 = F ~(x_1 - x_0).$$
Correction:
Work Energy is $F_{avg} \times d = \frac12 m(v_f^2-v_0^2)$
Impulse Momentum: $F_{avg} \times t = m(v_f−v_0)$
Which becomes,
$md(v_f-v_0)=\frac12 mt(v_f^2-v_0^2)$
masses cancel --->
$d(v_f-v_0)=\frac12 t(v_f^2-v_0^2)$
For a splat event where $v_f=0$, we get $-dv_0=-\frac12tv_0^2$
which simplifies to $v_0=2d/t$.
So, first question is can the two principles be combined in the first place as i've done. And then the second question is, if does everything reduce to $v_0=2d/t$ -- assuming a "splat" condition.
