User:Keand64

I am cetera.

If we assume a function that defines speed with resprect to time as so:

$$s(t) = jt^n + k,\!$$

we can derive

$$\begin{align} \int s(t) \, dt & = \int jt^n + k \, dt \\ & = \int jt^n \, dt + \int k \, dt \\ & = j\int t^n \, dt + k \int \, dt \\ & = j\frac{t^{n+1}}{n+1} + kt \\ \end{align}$$

thus obtaining a function that relates distance to time

$$d(t) = j\frac{t^{n+1}}{n+1} + kt,\!$$

likewise, we could derive from the original function

$$\begin{align} \frac{ds}{dt} & = \frac{d}{dt} jt^n + k \\ & = \frac{d}{dt} jt^n + \frac{d}{dt} k \\ & = j\frac{d}{dt} t^n + 0 \\ & = jnt^{n-1} \\ \end{align}$$

to obtain a function to relate acceleration to time

$$a(t) = jnt^(n-1),\!$$

In summary, at $$t,\!$$ seconds in time, the acceleration will be $$a(t),\!$$, the velocity will be $$s(t),\!$$, and the distance traveled will be $$d(t),\!$$.

An example:

$$\begin{align} j & = 1 \\ k & = 0 \\ n & = 2 \\ a(t) & = 2t \\ s(t) & = t^2 \\ d(t) & = \frac{t^3}{3} \\ \end{align}$$

so at five seconds after the initial time, the entity will have an acceleration of ten units of speed per a unit of time, a speed of twenty-five units of one-dimensional space per a unit of time, and will have travelled a total distance of about fourty-one units of one-dimensional space. (Exactly fourty-one and two thirds).