Here’s my version of the derivation of the kinetic energy formula (first post using latex, so go easy)
Kinetic energy is force times change in distance.
[tex]
KE = F \Delta x
[/tex]
Force is mass times acceleration, which is a change in velocity over time. We can write this as
[tex]
F= m \frac{\Delta v}{\Delta t}
[/tex]
Now the change in distance, [tex] \Delta x [/tex] can be expressed as velocity times time
[tex]
\Delta x = \bar{v} t
[/tex]
Notice that the velocity is is average velocity, where we have a constant change in velocity, or acceleration.
Average velocity can be written as the sum of the initial and final velocities divided by two.
[tex]
\bar v = \frac{v_0 + v_f}{2}
[/tex]
Now let's look at final velocity. We can write this as the initial velocity plus the change in velocity over time multiplied by time
[tex]
v_f = v_0 + \frac{\Delta v}{\Delta t} \Delta t
[/tex]
which can also be written as
[tex]
v_f = v_0 + at
[/tex]
Plugging this into the aveage velocity formula gives
[tex]
\bar v = \frac{v_0 + \left[ v_0 + \frac{\Delta v}{\Delta t} \Delta t \right]}{2}
[/tex]
which simplifies to
[tex]
\bar v = v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t
[/tex]
Now, we plug this into the change in distance formula
[tex]
\Delta x = \left[v_0 + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \right] \Delta t
[/tex]
[tex]
= v_0 \Delta t + \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t^2
[/tex]
Now, let's assume the beginning velocity is zero. Then we take the product of the above and force
[tex]
F\Delta x = \left( m\frac{\Delta v}{\Delta t} \right) \left( \frac{1}{2} \frac{\Delta v}{\Delta t} \Delta t \Delta t \right)
[/tex]
Collecting terms we arrive at
[tex]
KE = m \frac{1}{2} \Delta v \Delta v
[/tex]
Which of course is
[tex]
KE = \frac{1}{2} mv^2
[/tex]
