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Step-by-Step Solution
Step 1: Recognize the nature of uniform circular motion
In uniform circular motion, a particle travels in a circle at a constant speed $v$.
Despite the constant speed, the velocity vector changes its direction continuously,
giving rise to an acceleration — the centripetal acceleration — which always points
toward the center of the circle.
Step 2: Write the magnitude of centripetal acceleration
The magnitude of centripetal acceleration is
$$
a_c = \frac{v^2}{R},
$$
where $R$ is the radius of the circular path and $v$ is the speed of the particle.
Step 3: Determine the direction of the acceleration
Since the particle is at a point $P(R, \theta)$ on the circle, its position vector
from the origin (center of the circle) can be written as
$$
\vec{r} = R \cos\theta\, \hat{i} \;+\; R \sin\theta\, \hat{j}.
$$
The centripetal acceleration vector always points radially inward (i.e., it is in the
direction opposite to the position vector). Hence,
$$
\vec{a} = -\, a_c \frac{\vec{r}}{R}.
$$
Step 4: Express acceleration in component form
Substituting the magnitude $a_c = \frac{v^2}{R}$ and the position vector
$\vec{r} = (R\cos \theta)\hat{i} + (R\sin \theta)\hat{j}$ into
$\vec{a} = -\, a_c \frac{\vec{r}}{R}$, we get
$$
\vec{a} = -\, \frac{v^2}{R} \left(\cos \theta\, \hat{i} + \sin \theta\, \hat{j}\right).
$$
Simplifying,
$$
\vec{a} = -\, \frac{v^2}{R}\cos \theta \,\hat{i} \;-\; \frac{v^2}{R}\sin \theta \,\hat{j}.
$$
Step 5: State the final answer
Therefore, the acceleration of the particle in uniform circular motion at
the point $(R,\theta)$ is
$$
\boxed{- \frac{v^2}{R}\cos \theta \,\hat{i} - \frac{v^2}{R}\sin \theta \,\hat{j}}.
$$
