[Q1] The total acceleration vector $\vec{a}$ of the particle is the vector sum of its tangential component $\vec{a_t}$ and its radial component $\vec{a_r}$.

$$ \vec{a} = \vec{a_t} + \vec{a_r} $$

The velocity vector $\vec{v}$ is always tangent to the circular path, so it is in the same direction as the tangential acceleration $\vec{a_t}$. Therefore, the angle $\alpha$ between $\vec{a}$ and $\vec{v}$ is the same as the angle between $\vec{a}$ and $\vec{a_t}$.

The tangential and radial acceleration vectors are perpendicular. From the vector diagram, we can relate their magnitudes:

$$ \tan(\alpha) = \frac{|\vec{a_r}|}{|\vec{a_t}|} = \frac{a_r}{a_t} $$

For uniformly accelerated circular motion with radius $R$ and constant angular acceleration $\beta$, the components of acceleration are:

1. Tangential acceleration: $a_t = R\beta$ (constant).
2. Radial (centripetal) acceleration: $a_r = \omega^2 R$, where $\omega$ is the instantaneous angular velocity.

The particle starts from rest, so its initial angular velocity is $\omega_0 = 0$. The rotational kinematic equation relating angular velocity $\omega$ and angular displacement (the central angle $\theta$) is:

$$ \omega^2 = \omega_0^2 + 2\beta\theta = 2\beta\theta $$

Note that $\theta$ must be in radians.

Now we can express the radial acceleration in terms of $\theta$:

$$ a_r = (2\beta\theta)R $$

Finally, substitute the expressions for $a_t$ and $a_r$ into the equation for $\tan(\alpha)$:

$$ \tan(\alpha) = \frac{a_r}{a_t} = \frac{(2\beta\theta)R}{R\beta} $$

Simplifying the expression, we find the relationship between $\alpha$ and $\theta$:

$$ \tan(\alpha) = 2\theta $$