Question

Prove v^2=u^2+2as using the methods of calculus.

Okay, I did it in one way and I showed it to my teacher but she said that there was another way, besides the method which I used. Can someone show me that other way?

Answer 1

My method was this;

\[a=\frac{ dv }{ dt }=\frac{ dv }{ ds }\frac{ ds }{ dt} \rightarrow \int\limits_{_{x0}}^{x}a ds=\int\limits_{_{v0}}^{v} v dv \rightarrow a \Delta x =\frac{ 1 }{ 2 }(v ^{2}-v _{0}^{2})\]

And then, I can obtain the equation I was looking for. But is there another way to derive it using the methods of calculus?

Answer 2

From the Conservation of Energy:

The initial kinetic energy of the particle plus some extra energy,
will be equal to the final energy the particle.
\[KE_0+EE=KE_f\]
Using the formula for kinetic energy \(KE=\tfrac12mv^2\),
[where the initial velocity is \(u\), and the final velocity is \(v\)]

The extra energy to the particle, does work \(EE=W=Fs\),
Force is \(F=ma\)

\[\frac12mu^2+mas=\frac12mv^2\\\text{(multiplying by $2$, dividing my $m$)}\\
\qquad \ u^2+2as=v^2\]

Answer 3

(whoops, i didn't use calculus)