I know this formula $D = vt + \frac{1}{2}at^2$ for calculating the distance given initial velocity, time and acceleration. But what if my acceleration is not static, but increasing exponentially defined by $f(n) = a^n$? If the acceleration $a$ starts increasing at distance $x$ and time $t_x$ (reference point a) and I do a measurement after $n$ seconds (reference point b) with the formula above, the results would be incorrect, because the acceleration initially was smaller.
Example: at reference point a object travels at velocity 10 m/s and acceleration is $2 \frac{m}{s^2}$ and time is 40 seconds ($t_x$). After this point acceleration starts to increase by $a(n) = a^n$ where $n$ is every second after $t_x$. So at $n=2$ (total 42 seconds) seconds acceleration would be $4 \frac{m}{s^2}$, $n=3$ it would be $8 \frac{m}{s^2}$ and so on. I want to measure the distance from $a$ to $b$, where $b$ is after $n=10$ (total 50 seconds).
Example2: at reference point a object travels at velocity 10 m/s and acceleration is $2 \frac{m}{s^2}$ and time is 40 seconds ($t_x$). After this point acceleration starts to increase by $a(n) = a^dn$ where $dn$ is every meter after a. So at $dn=2$ (total $a+2$ meters) seconds acceleration would be $4 \frac{m}{s^2}$, $dn=3$ it would be $8 \frac{m}{s^2}$ and so on. I want to measure the distance from a to b, where b is after after 10 seconds (total 50 seconds).