If I have initial velocity $V_0$ and I want to know how much distance is covered after $t$ seconds with acceleration $a$ meters per second, I tried this (where $V_t = V_0 + at$):
$$d = \lim_{k\to\infty}\frac{t}{k} \sum_{i=0}^{k-1} \left(V_0 + \frac{i \cdot (V_t-V_0)}{k-1}\right)$$
For example if $V_0 = 20$ and $t=3$ and $a=10$ then we could keep approximating:
$V_3 = \frac{3}{2}(20) + \frac{3}{2}(50)$
$V_3 = \frac{3}{3}(20) + \frac{3}{3}(35) + \frac{3}{3}(50)$
$V_3 = \frac{3}{4}(20) + \frac{3}{4}(30) + \frac{3}{4}(40) +\frac{3}{4}(50)$
And so on. I generalized it to $k$ terms and went to infinity and simplified the algebra. If you do the math the $k$ terms cancel out and you get this:
$V_t = V_0t + \frac{1}{2}at^2$
Which seems to be correct from what I am see on Google.
But how do I make the connection to calculus where people start talking about derivatives and integrals regarding distance, velocity, and acceleration? I don't understand how we "know" to use those things there and why they match the result I just got.
$d(t) = \int_0^t v(\tau) d \tau = \int_0^t (v_0+ a\tau) d \tau$.
Evaluating the Riemann integral using a uniform grid gives $d(t) = \lim_k {t \over k} \sum_{i=0}^{k-1} V_{{i t \over k}} = \lim_k {t \over k} \sum_{i=0}^{k-1} (V_0 + a{i t \over k}) = V_0 t + {1 \over 2} a t^2$.