1. Position vector of the particle

For a point $P(R,\theta)$ on a circle of radius $R$, measured from the $x$-axis,

$\vec r = R\cos\theta\,\hat i + R\sin\theta\,\hat j$

2. Direction of acceleration in uniform circular motion

In uniform circular motion, the speed is constant, but the velocity direction changes continuously.

Hence the acceleration is centripetal, i.e. directed towards the center.

Its magnitude is

$a = \frac{v^2}{R}$

and its direction is opposite to the radial unit vector.

3. Radial unit vector

The outward radial unit vector at angle $\theta$ is

$\hat r = \cos\theta\,\hat i + \sin\theta\,\hat j$

Therefore, inward direction is

$-\hat r = -\cos\theta\,\hat i - \sin\theta\,\hat j$

4. Acceleration vector

Thus,

$\vec a = \frac{v^2}{R}(-\hat r)$

$\vec a = -\frac{v^2}{R}\cos\theta\,\hat i - \frac{v^2}{R}\sin\theta\,\hat j$

5. Compare with options

• A: $-\dfrac{v^2}{R}\cos\theta\,\hat i + \dfrac{v^2}{R}\sin\theta\,\hat j$ ❌
• B: $-\dfrac{v^2}{R}\sin\theta\,\hat i + \dfrac{v^2}{R}\cos\theta\,\hat j$ ❌
• C: $-\dfrac{v^2}{R}\cos\theta\,\hat i - \dfrac{v^2}{R}\sin\theta\,\hat j$ ✅
• D: $\dfrac{v^2}{R}\hat i + \dfrac{v^2}{R}\hat j$ ❌

Therefore, the correct option is C.