# Properly deriving the formula for the kinetic energy

I am a physics student and recently attended my first lectures. One thing that strikes me even now was the way the professor derived the formula for the kinetic energy.

He started with the (already given) definition of "work": $$W = \int_{p_{start}}^{p_{end}} F(s) \space ds = \int_{p_{start}}^{p_{end}} m \space a(s) \space ds.$$

Then, he did this: $$W = \int_{t_{start}}^{t_{end}} m \space a \space v \space dt$$ Without explaining, what exactly he did. I am assuming this was some kind of "Integration by substitution" due to $\frac{ds}{dt} = v$ and thus $ds = v \space dt$. The question that bugs me is what happened to the $t$ in $a(t)$?

He continued with $$W = \int_{t_{start}}^{t_{end}} m \space \frac{dv}{dt} \space v \space dt$$ and cancelled out the $dt$s: $$W = \int_{v_{start}}^{v_{end}} m \space v \space dv$$ What exactly happened here?

The rest was simple integration. But I cannot see what exactly happened in both of theses steps. Can someone explain this to me and give me a mathematically correct alternative?

• $a(t)=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$, with $\frac{dx}{dt}=v$ you get $dW=ma dx=mv dv$. Note I used $x$ instead of $s$.
– Gert
Commented Nov 4, 2015 at 21:19
• @Bill - Maybe he thinks that he will get a clearer explanation from us.
– user93237
Commented Nov 4, 2015 at 21:21
• Hint: See the work-energy theorem and Wikipedia. Commented Nov 4, 2015 at 21:25
• "The question that bugs me is what happened to the t in a(t)?" - Do you mean what happened to the 's' in a(s)? When he switched from doing an integration over 's' to an integration over 't', he had to switch the function 'a' from being a function of the distance 's' to a function of the time 't'.
– user93237
Commented Nov 4, 2015 at 21:27
• BTW, is there a guide on this site for how to format mathematical expressions? Are you guys using LaTex or something like that?
– user93237
Commented Nov 4, 2015 at 21:31

$$W=\int_{x1}^{x2}ma(s)ds$$ Since $ds/dt=v\Leftrightarrow ds=vdt$: $$W=\int_{t1}^{t2}ma(t)vdt$$ The expression for acceleration should of course be $a(t)$, since it might depend on $t$ (just as it depends on $s$). But writing it as just $a$ is not "wrong" (the $(t)$ just emphasizes the dependency on $t$, and $a=a(t)$ still counts). To keep it consistent through the derivation, I do though agree on it being pedagogically smarter to use $a(t)$.
Since $a=dv/dt$: $$W=\int_{t1}^{t2}m\frac{dv}{dt}vdt$$ Remember here what the differentials originally mean: $dt$ is an infinitely small change in time and $dv$ is an infinitely small change in speed. Dividing the variable $dt$ with itself gives $1$. So the next step is simply a reduction of the expression: $$W=\int_{v1}^{v2}mvdv$$
A note: Changing the integration limits from $x$ to $t$ to $v$ doesn't matter, as long as we still integrate over the same part. So as long as $x1$ is the value at $t1$ and the speed here is $v1$ (and same for the end-point), then the path integrated over is the same.