So if calculating the change in an object's position (with a constant acceleration) is done with this equation:
$o = vt + (\frac12)a t^2$
$o$ is offset from original position
$v$ is starting velocity
$t$ is time
$a$ is acceleration
How would I calculate change in position if acceleration is changing (at a fixed rate).
Like thus?
$o = vt + (\frac12)a t^2 + (\frac12)i t^{2^2}$
or
$o = vt + (\frac12)a t^2 + (\frac12)i t^3$
$i$ is increase rate of acceleration
Best Answer
Hint:
From the definition of acceleration $a$ and velocity $v$ you know that $$ a=\frac{dv}{dt}=\frac{d}{dt}\frac{dx}{dt}=\frac{d^2x}{dt^2} $$ where $x$ is the position (the space). So, if $a=it$ you have to solve the equation: $$ \frac{d^2x}{dt^2}= it $$
can you do this?
If you don't know the calculus (derivatives and integrals) it is difficult to find the equation of motion for non constant acceleration.
In your case, if $a=it$ and $i$ is a constant, we can find the velocity $v(t)$ in a way analogous to that used in the uniformly accelerated motion for to find the position $x(t)$ from the velocity $v=at$, and this gives: $v(t)=\frac{1}{2}it^2+v_0$.
But now we have to find the position $x(t)$ and this cannot be done in a simple way without the use of calculus.
The idea is that we start from: $$ \frac{dx}{dt}=v(t)=\frac{1}{2}it^2+v_0 $$ Where $dx$ is the little (infinitesimal) change of position for an infinitesimal time interval $dt$, so that the infintesimal change of position $dx$ is: $$ dx=\left(\frac{1}{2}it^2+v_0 \right)dt $$ than we sum up all this infinitesimal displacements to find the position at time $t$. This intuition can be defined rigorously and captured in the notion of integral.
The final result , in your case, is the equation of motion: $$ x(t)=\frac{1}{6}it^3+v_0t+s_0 $$