- #1
FranzS
- 86
- 26
- TL;DR
- How to find an equation for acceleration vs. position in a constant jerk motion?
Hi PF,
a motion with constant jerk ##j##, i.e. with a constant rate of change of the acceleration with respect to time, should be described by the following equations (hope the variable names are self-explanatory):
$$
\begin{cases}
a = a_0 + j \ t \\
v = v_0 + a_0 \ t + \frac{1}{2} \ j \ t^2 \\
s = s_0 + v_0 \ t + \frac{1}{2} \ a_0 \ t^2 + \frac{1}{6} \ j \ t^3
\end{cases}
$$
I'm interested in deriving an explicit equation for acceleration ##a## with respect to position ##s## as the only independent variable. Jerk ##j## and all initial conditions are to be considered known values.
My guess is that I should set up a differential equation of the type...
$$
\frac{da(s)}{ds}=\frac{da(t)}{dt} \cdot \frac{dt(s)}{ds}
$$
... where ##da(t)/dt=j## by definition, and then integrate both sides.
But that would mean finding ##t(s)## first, and that's solving a cubic equation.
Is my approach correct? Is there any simpler way?
Thank you.
a motion with constant jerk ##j##, i.e. with a constant rate of change of the acceleration with respect to time, should be described by the following equations (hope the variable names are self-explanatory):
$$
\begin{cases}
a = a_0 + j \ t \\
v = v_0 + a_0 \ t + \frac{1}{2} \ j \ t^2 \\
s = s_0 + v_0 \ t + \frac{1}{2} \ a_0 \ t^2 + \frac{1}{6} \ j \ t^3
\end{cases}
$$
I'm interested in deriving an explicit equation for acceleration ##a## with respect to position ##s## as the only independent variable. Jerk ##j## and all initial conditions are to be considered known values.
My guess is that I should set up a differential equation of the type...
$$
\frac{da(s)}{ds}=\frac{da(t)}{dt} \cdot \frac{dt(s)}{ds}
$$
... where ##da(t)/dt=j## by definition, and then integrate both sides.
But that would mean finding ##t(s)## first, and that's solving a cubic equation.
Is my approach correct? Is there any simpler way?
Thank you.