In constant accelaration and linear (1D) motion, we can show that relationship between velocity and position is quadratic (parabola) by
We can write $v$ in the form of $v=v(x(t))$ \begin{equation} a=\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \\ a=v\cdot\frac{dv}{dx}\\ \int a\text{ }dx = \int v\cdot\frac{dv}{dx} \text{ } dx\\ ax = \frac{v^2}{2} + C \end{equation} This relationship is not complicated to prove, it is also does not involve time relationship between velocity or position (that is we do not need to find v(t) and x(t).) Then, I think that it can be done in similar way under the condition constant jerk(and of course not 0).
My Attempt: Jerk is defined by $J=\frac{da}{dt}$
\begin{equation} J=\frac{da}{dt}=\frac{da}{dv} \cdot \frac{dv}{dt}\\ J=a \cdot \frac{da}{dv} \\ Jv+C_1 = \frac{a^2}{2}\\ a= \pm\sqrt{2Jv+C_2}\text{ }; C_2 = 2C_1 \end{equation} Acceleration is defined by $a=\frac{dv}{dt}$ \begin{equation} a=\frac{dv}{dt}=\frac{dv}{dx} \cdot \frac{dx}{dt}\\ a=v \cdot \frac{dv}{dx}\\ \pm\sqrt{2Jv+C_2}=v \cdot \frac{dv}{dx} \end{equation} Solve this DE and we will get \begin{equation} x=\pm \frac{(Jv-C_2)\sqrt{2Jv+C_2}}{3J^2}+C_3 \end{equation} This is where I start to think if there are any mistakes in my solution. Because the solution seems to be very ugly. But if the solution is correct, I would like to see an easier proof. So please verify this proof.


You have $$J = \frac{{\rm d}a}{{\rm d}t} = \frac{{\rm d}a}{{\rm d}v} \frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}a}{{\rm d}v} a$$

With $v_1$ and $a_1$ as initial conditions then:

$$\left. \int J {\rm d}v = \int a {\rm d}a = \frac{1}{2} a^2 + C \\ J (v-v_1) = \frac{1}{2} \left( a^2 - a_1^2 \right) \right\} a = \sqrt{a_1^2+2 J (v-v_1)}$$

Position is found from acceleration with

$$ a=\frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} v $$

$$ x = \int \frac{v}{a} {\rm d}v = \int \frac{v}{\sqrt{a_1^2+2 J (v-v_1)}} {\rm d}v$$

$$ x - x_1 = \frac{ \left( J (v+2 v_1)-a_1^2 \right) \sqrt{2 J(v-v_1)+a_1^2} - a_1 (3J v_1-a_1^2)}{3 J^2} $$

So no, with constant jerk, you do not have a parabolic relationship between position and velocity.

There is cubic relationship though that is described as

$$ (x-c_1)^2 = \frac{v-c_2}{c_4} + \frac{(v-c_3)^3}{c_5} $$

$$ \begin{aligned} c_1 & = \frac{3 J^2 x_1 -3 J a_1 v_1 + a_1^3}{3 J^2} \\ c_2 & = \frac{5 (a_1^2-2 J v_1)}{6 J} \\ c_3 & = \frac{a_1^2-2 J v_1}{2 J} \\ c_4 & = \frac{6 J^3}{a_1^2 (4 J v_1 - a_1^2)-4 J^2 v_1^2} \\ c_6 & = \frac{9 J}{2} \end{aligned} $$


Let me rephrase your first part, with constant $a$. The standard solutions, involving time, are: $$a(t)=a\\ v(t)=a t +v_0\\ x(t)=\frac{1}{2}a t^2+v_0 t+x_0$$

From this, it is clear how to get your solution. Eliminate $t$ from the last 2 equations, e.g. by completion of squares: $$a x(t)=\frac{1}{2}[a^2 t^2+2 a v_0 t+v_0^2]+[a x_0-\frac{1}{2}v_0^2]=\frac{1}{2} v(t)^2+C $$ You can read off your constant $C=[a x_0-\frac{1}{2}v_0^2]$.

Generalizing this to constant "jerk" $J$, you will get: $$J(t)=J\\a(t)=J t+a_0\\ v(t)=\frac{1}{2}J t^2+a_0 t +v_0\\ x(t)=\frac{1}{6}J t^3+\frac{1}{2}a_0 t^2+v_0 t+x_0$$

In an analogous way as above you find: $J v(t)=\frac{1}{2} a(t)^2+C'$ where $C'=[J v_0-\frac{1}{2}a_0^2]$ relates to your $C_1$.

Finally, you have two options to obtain $t$:

(i) from the $a(t)$ equation: this gives a linear $t(a)$ which you can plug in $x(t)$ to obtain $x(a)$, and everything goes fine in the $J\to 0$ limit;

(ii) from the quadratic $v(t)$ equation, which gives you the $\pm\sqrt{\ldots}$ stuff, and which goes weird in the $J\to 0$ limit.


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