Non-uniform circular motion with constant radius of curvature

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Suppose we have a car moving on a circular track of radius $$b$$ and speed $$v=ct$$, where $$t$$ is time and $$c\in\mathbb{R^+}$$.

Let the velocity vector be written as: $$\vec v = v\vec u$$ where $$|\vec v|=v=ct$$ and $$\vec u$$ is a unit vector in the direction of motion. Thus by the Chain Rule, $$\vec{\dot v}=\vec{a}=v'\vec u+v\vec{\dot u}\tag1$$

But in general for nonuniform motion in a circle, $$\vec{a}=a_T\vec{u}+a_R\vec r\tag2$$ where $$\vec{r}$$ is a unit radial vector always pointing inwards to the center of the circle, $$a_T$$ is the magnitude of the tangential acceleration in the direction of $$\vec v$$, and $$a_R$$ is the magnitude of the radial acceleration.

Reconciling $$(1)$$ and $$(2)$$, one finds $$a_T=v'$$ but its not so clear for $$a_R$$. I would like to somehow equate $$v\vec{\dot u }=a_R\vec r$$ by using the more general formula that $$a_R=\frac{v^2}{r}$$ But for this particular problem $$v\vec{\dot u}=ct\vec{\dot u}=^? \frac{c^2t^2}{b}\vec r$$ So it must be then that since $$\vec{u}$$ and $$\vec{\dot u}$$ are perpendicular that $$\frac{t}{b}\vec{\dot u}=\vec r$$ right? How can I proceed further?

I'm hoping to utilize the above work to help me show the $$\theta$$ such that $$\vec v\cdot \vec a=va\cos\theta$$ will be equal to $$45^{\circ}$$ at $$t=\sqrt{\frac{b}{c}}$$.

You had the correct tangential, $$c$$, and radial, $$\dfrac {c^2t^2}{b}$$, accelerations
If the radius is constant then in polar coordinates $$\vec v = b\,\dot \theta \hat \theta = c\,t\, \hat \theta \Rightarrow \dot \theta = \dfrac{ct}{b} \Rightarrow \ddot \theta = \dfrac cb$$ and $$\vec a = -b\,\dot \theta^2\, \hat r + b\,\ddot\theta \,\hat \theta =-\dfrac{c^2t^2}{b}\hat r +c\,\hat \theta = - c\,\hat r + c\,\hat \theta$$ at $$t=\sqrt{\dfrac bc}$$ so the angle between $$\vec v$$ and $$\vec a$$ is $$45^\circ$$.