Derive the equations of motion for constant acceleration using the method of calculus.

(N/A) By definition,acceleration is $a = \frac{dv}{dt}$.
Rearranging gives $dv = a dt$.
Integrating both sides from initial velocity $v_0$ at $t=0$ to $v$ at time $t$:
$\int_{v_0}^{v} dv = \int_{0}^{t} a dt = a \int_{0}^{t} dt$
$v - v_0 = at \implies v = v_0 + at$.
Since $v = \frac{dx}{dt}$,we have $dx = v dt = (v_0 + at) dt$.
Integrating both sides from initial position $x_0$ at $t=0$ to $x$ at time $t$:
$\int_{x_0}^{x} dx = \int_{0}^{t} (v_0 + at) dt$
$x - x_0 = v_0 t + \frac{1}{2} at^2 \implies x = x_0 + v_0 t + \frac{1}{2} at^2$.
Using the chain rule,$a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$.
Thus,$v dv = a dx$.
Integrating both sides from $v_0$ to $v$ and $x_0$ to $x$:
$\int_{v_0}^{v} v dv = \int_{x_0}^{x} a dx$
$\frac{v^2 - v_0^2}{2} = a(x - x_0)$
$v^2 = v_0^2 + 2a(x - x_0)$.