Why $F = m(v_f - v_0)/2$? Force is directly proportional to mass and velocity and inversely proportional to time so why don't we write $F=1/t+m+v-v_0$ where $m$ is mass, $v$ is final velocity, and $v_0$ is initial velocity?
 A: 
Force is directly proportional to mass and velocity and inversely proportional to time so why don't we write $F=1/t+m+v-v_0$

As others mentioned, the units don’t work. However, suppose we modify it to $$F=k_t/t+k_m m+k_v(v-v_0)$$ where the various $k$ are constants with appropriate dimensions that make each term a force.
Now, you have an equation that is dimensionally consistent. However, it is not directly proportional to mass and velocity and inversely proportional to time. If you double $m$ then according your formula you do not double $F$. Instead you get $$k_t/t+k_m 2m+k_v(v-v_0) \ne 2 F$$
For $F$ to be proportional to $m$ means $F=km$, and similarly with the other factors.
Also, one nitpick. Force is not proportional to velocity but the average force is proportional to the change in velocity. Those are slightly different statements.
A: One way to convince yourself of this is that force is measured in $\rm{kg\cdot m/s^2}$ which is a Newton. This can be formed by multiplying mass kg times velocity m/s divided by time s.  You cannot get there by adding the units.
More generally, you cannot add any units (or the corresponding dimensioned quantities) unless they are the same. You can add two masses
1 kg + 2 kg = 3 kg
Or two velocities
2 m/s + 4 m/s = 6 m/s
Or two forces
1 $\rm{kg\cdot m/s^2}$ + 0.5 $\rm{kg\cdot m/s^2}$ = 1.5 $\rm{kg\cdot m/s^2}$
But you cannot add a mass and a length
1 kg + 2 meters =.....
It doesn't work.
A: By Newton's second law force is defined as mass times acceleration.  Acceleration is defined as the time derivative of velocity, and velocity is defined as the time derivative of position.  Force, acceleration, velocity, and position are vectors.
