Calculo e derivação baseados em 1 dimensão: MUV

If you know calculus, you may quickly derive the most common equations of motion.

Suppose we have a constant acceleration $a$, an initial velocity $v_i$, and an initial position $x_i$. Denote the final velocity and final position as $v_f$ and $x_f$ respectively, and let $t_i$ and $t_f$ be the initial and final time, respectively.

From calculus, we know: $$\frac{dv}{dt}=a$$ Integrating, $$dv=a\cdot dt$$ $$\int_{t_i}^{t_f}dv=\int_{t_i}^{t_f}a\cdot dt$$ $$[v]_{t_i}^{t_f}=[at]_{t_i}^{t_f}$$ $$v_f-v_i=a(t_f-t_i)$$ If we let the change in time equal $t$, then we get the familiar equation: $$v_f=v_i+at$$

Now, we may apply the same process to solve for the position: $$\frac{dx}{dt}=v$$ $$dx=v\cdot dt$$ $$\int_{t_i}^{t_f}dx=\int_{t_i}^{t_f}v\cdot dt$$ $$\int_{t_i}^{t_f}dx=\int_{t_i}^{t_f}(v_i+a(t-t_i))\cdot dt$$ $$[x]_{t_i}^{t_f}=[v_it+\frac{1}{2}a(t-t_i)^2]_{t_i}^{t_f}$$ $$x_f-x_i=v_i(t_f-t_i)+\frac{1}{2}a(t_f-t_i)^2+0$$ If we let the change in time equal $t$, then we get the familiar equation: $$x_f=x_i+v_it+\frac{1}{2}at^2$$

Now, let us evaluate this integral: $$\int_{v_i}^{v_f}2v\cdot dv$$ A simple integration yields the result $v_f^2-v_i^2$.

On the other hand, we could evaluate this integral by substituting $t$ as our variable instead of $v$.

Our new integral is: $$\int_{t_i}^{t_f}2v(a\cdot dt)$$ Clearly $2a$ is a constant, so we may move it out from the integral: $$2a\int_{t_i}^{t_f}v\cdot dt$$ But that integral is simply the displacement, $x_f-x_i$! Let us call this $x$. Then, equating the two expressions: $$v_f^2-v_i^2=2ax$$ This is our familiar time-excluding kinematics equation. Keep it in mind if you are not given time, but you are given much of the other information.