A particle moves in a circle of radius R. In half the period of revolution its displacement is ____ and distance covered is ____.

Step 1: Identify the given parameters and the time interval.

The particle moves in a circle of radius $R$ with a time period of revolution $T$. The time interval of interest is half the period:

$$t = \frac{T}{2}$$

Step 2: Calculate the angular displacement.

The angular velocity $\omega$ is related to the time period $T$ by:

$$\omega = \frac{2\pi}{T}$$

In half the period, the angular displacement $\theta$ is:

$$\theta = \omega \cdot t$$ $$\theta = \frac{2\pi}{T} \cdot \frac{T}{2} = \pi$$

An angular displacement of $\pi$ radians ($180^\circ$) means the particle moves from its starting point $A$ to the diametrically opposite point $B$.

Step 3: Calculate the distance covered.

The distance covered is the actual path length, which is the arc length of the semi-circle.

$$\text{Distance} = R\theta$$ $$\text{Distance} = \pi R$$

Step 4: Calculate the displacement.

The displacement is the straight-line distance between the initial position $A$ and the final position $B$, which corresponds to the diameter of the circle.

$$\text{Displacement} = 2R$$

Answer: Displacement is $2R$ and distance covered is $\pi R$.