With an aid of differential geometry, velocity, acceleration and jerk can be written as:

\begin{align*} \mathbf{v} &= \dot{s} \, \mathbf{T} \\ &= v \, \mathbf{T} \\ \mathbf{a} &= \ddot{s} \, \mathbf{T}+ \kappa \, \dot{s}^2\mathbf{N} \\ \mathbf{b} &= (\dddot{s}-\kappa^2 \dot{s}^3) \mathbf{T}+ (3\kappa \dot{s} \ddot{s}+\dot{\kappa} \dot{s}^2) \mathbf{N}+ \kappa \tau \dot{s}^3 \mathbf{B} \\ \end{align*}

Now $$ \mathbf{a}=\boldsymbol{\omega} \times \mathbf{v} \implies \mathbf{a} \perp \mathbf{T} \implies \ddot{s}=0 \implies \dot{s}=\text{constant} \implies v=u$$

and $$ \mathbf{b}= \dot{\mathbf{a}}=\boldsymbol{\omega} \times \mathbf{a} \implies \mathbf{b} \perp \mathbf{a} \implies \mathbf{b} \perp \mathbf{N} \implies \dot{\kappa}\dot{s}^2=0 \implies \kappa=\text{constant}$$

Also, \begin{align*} \int \mathbf{a} \, dt &= \boldsymbol{\omega} \times \int \mathbf{v} \, dt \\ \mathbf{v} &= \mathbf{u}+\boldsymbol{\omega} \times \mathbf{r} \\ \mathbf{r}(0) &= \mathbf{0} \\ \dot{\mathbf{r}}(0) &= \mathbf{u} \\ \boldsymbol{\omega} \cdot \mathbf{v} &= \boldsymbol{\omega} \cdot \mathbf{u} \\ &= \text{constant} \end{align*}

We have

\begin{align*} \mathbf{v} \times \mathbf{a} &= \mathbf{v} \times (\boldsymbol{\omega} \times \mathbf{v}) \\ &= v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v} \\ \kappa &= \frac{|v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v}|} {v^3} \\ &= \frac{|\boldsymbol{\omega} \times \mathbf{v}|}{v^2} \\ &= \frac{|\boldsymbol{\omega} \times \mathbf{u}|}{u^2} \\ |\mathbf{a}| &= |\boldsymbol{\omega} \times \mathbf{u}| \\ &= \text{constant} \\ \mathbf{a} \times \mathbf{b} &= a^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{a}) \mathbf{a} \\ &= a^2 \boldsymbol{\omega} \\ \tau &= \frac{\mathbf{v} \cdot a^2\boldsymbol{\omega}} {(\mathbf{v} \times \mathbf{a})^2} \\ &= \frac{\boldsymbol{\omega} \cdot \mathbf{v}}{v^2} \\ &= \frac{\boldsymbol{\omega} \cdot \mathbf{u}}{u^2} \\ &= \text{constant} \end{align*}

Both $\kappa$ and $\tau$ are constants implying the path is helical. If $\mathbf{u} \cdot \boldsymbol{\omega}=0 \implies \tau=0$, then it'll be a circle. While $\mathbf{u} \times \boldsymbol{\omega}=\mathbf{0} \implies \kappa=0$, that'll be a straight line.

Fitting with initial conditions:

$$\fbox{$\quad \mathbf{r}=\mathbf{u}t+ \frac{\mathbf{u} \times \boldsymbol{\omega}}{\omega^2}(\cos \omega t-1)+ \frac{\boldsymbol{\omega} \times (\mathbf{u} \times \boldsymbol{\omega})}{\omega^{3}}(\sin \omega t-\omega t) \quad \\$}$$

Some facts from differential geometry \begin{align*} s &= \int |\mathbf{v}| \, dt \tag{arclength} \\ \dot{s} &= |\mathbf{v}| \tag{speed} \\ &= v \\ \mathbf{T} &= \frac{\mathbf{v}}{v} \tag{tangent vector}\\ \mathbf{B} &= \frac{\mathbf{v} \times \mathbf{a}}{|\mathbf{v} \times \mathbf{a}|} \tag{binormal vector} \\ \mathbf{N} &= \mathbf{B} \times \mathbf{T} \tag{normal vector} \\ \kappa &= \frac{|\mathbf{v} \times \mathbf{a}|}{v^3} \tag{curvature} \\ \tau &= \frac{\mathbf{v} \cdot \mathbf{a} \times \mathbf{b}} {(\mathbf{v} \times \mathbf{a})^2} \tag{torsion} \end{align*}

With an aid of differential geometry, velocity, acceleration and jerk can be written as:

\begin{align*} \mathbf{v} &= \dot{s} \, \mathbf{T} \\ &= v \, \mathbf{T} \\ \mathbf{a} &= \ddot{s} \, \mathbf{T}+ \kappa \, \dot{s}^2\mathbf{N} \\ \mathbf{b} &= (\dddot{s}-\kappa^2 \dot{s}^3) \mathbf{T}+ (3\kappa \dot{s} \ddot{s}+\dot{\kappa} \dot{s}^2) \mathbf{N}+ \kappa \tau \dot{s}^3 \mathbf{B} \\ \end{align*}

Now $$ \mathbf{a}=\boldsymbol{\omega} \times \mathbf{v} \implies \mathbf{a} \perp \mathbf{T} \implies \ddot{s}=0 \implies \dot{s}=\text{constant} \implies v=u$$

and $$ \mathbf{b}= \dot{\mathbf{a}}=\boldsymbol{\omega} \times \mathbf{a} \implies \mathbf{b} \perp \mathbf{a} \implies \mathbf{b} \perp \mathbf{N} \implies \dot{\kappa}\dot{s}^2=0 \implies \kappa=\text{constant}$$

Also, \begin{align*} \int \mathbf{a} \, dt &= \boldsymbol{\omega} \times \int \mathbf{v} \, dt \\ \mathbf{v} &= \mathbf{u}+\boldsymbol{\omega} \times \mathbf{r} \\ \mathbf{r}(0) &= \mathbf{0} \\ \dot{\mathbf{r}}(0) &= \mathbf{u} \\ \boldsymbol{\omega} \cdot \mathbf{v} &= \boldsymbol{\omega} \cdot \mathbf{u} \\ &= \text{constant} \end{align*}

We have

\begin{align*} \mathbf{v} \times \mathbf{a} &= \mathbf{v} \times (\boldsymbol{\omega} \times \mathbf{v}) \\ &= v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v} \\ \kappa &= \frac{|v^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{v})\mathbf{v}|} {v^3} \\ &= \frac{|\boldsymbol{\omega} \times \mathbf{v}|}{v^2} \\ &= \frac{|\boldsymbol{\omega} \times \mathbf{u}|}{u^2} \\ |\mathbf{a}| &= |\boldsymbol{\omega} \times \mathbf{u}| \\ &= \text{constant} \\ \mathbf{a} \times \mathbf{b} &= a^2 \boldsymbol{\omega}-(\boldsymbol{\omega} \cdot \mathbf{a}) \mathbf{a} \\ &= a^2 \boldsymbol{\omega} \\ \tau &= \frac{\mathbf{v} \cdot a^2\boldsymbol{\omega}} {(\mathbf{v} \times \mathbf{a})^2} \\ &= \frac{\boldsymbol{\omega} \cdot \mathbf{v}}{v^2} \\ &= \frac{\boldsymbol{\omega} \cdot \mathbf{u}}{u^2} \\ &= \text{constant} \end{align*}

Both $\kappa$ and $\tau$ are constants implying the path is helical. If $\mathbf{u} \cdot \boldsymbol{\omega}=0 \implies \tau=0$, then it'll be a circle. While $\mathbf{u} \times \boldsymbol{\omega}=\mathbf{0} \implies \kappa=0$, that'll be a straight line.

Fitting with initial conditions:

$$\fbox{$\quad \mathbf{r}=\mathbf{u}t+ \frac{\mathbf{u} \times \boldsymbol{\omega}}{\omega^2}(\cos \omega t-1)+ \frac{\boldsymbol{\omega} \times (\mathbf{u} \times \boldsymbol{\omega})}{\omega^{3}}(\sin \omega t-\omega t) \quad \\$}$$

Some facts from differential geometry \begin{align*} s &= \int |\mathbf{v}| \, dt \tag{arclength} \\ \dot{s} &= |\mathbf{v}| \tag{speed} \\ &= v \\ \mathbf{T} &= \frac{\mathbf{v}}{v} \tag{tangent vector}\\ \mathbf{B} &= \frac{\mathbf{v} \times \mathbf{a}}{|\mathbf{v} \times \mathbf{a}|} \tag{binormal vector} \\ \mathbf{N} &= \mathbf{B} \times \mathbf{T} \tag{normal vector} \\ \kappa &= \frac{|\mathbf{v} \times \mathbf{a}|}{v^3} \tag{curvature} \\ \tau &= \frac{\mathbf{v} \cdot \mathbf{a} \times \mathbf{b}} {(\mathbf{v} \times \mathbf{a})^2} \tag{torsion} \end{align*}