The velocity-time plot of a particle travelling along a straight...

From the given velocity-time graph, the curve is a straight line. The slope of a $v-t$ graph represents acceleration. Since the slope is constant, the particle exhibits a constant acceleration. Thus, option (1) is correct.

The distance covered in a time interval is the magnitude of the area under the $v-t$ graph.
For the interval $0$ to $10 \text{ s}$:
Distance $d_1 = \dfrac{1}{2} \times 10 \times 10 = 50 \text{ m}$
Average speed $= \dfrac{d_1}{\Delta t} = \dfrac{50}{10} = 5 \text{ m/s}$

For the interval $10 \text{ s}$ to $20 \text{ s}$:
From the graph's constant slope, at $t = 20 \text{ s}$, the velocity is $v = -10 \text{ m/s}$.
Distance $d_2 = \dfrac{1}{2} \times (20 - 10) \times |-10| = 50 \text{ m}$
Average speed $= \dfrac{d_2}{\Delta t} = \dfrac{50}{10} = 5 \text{ m/s}$
The average speeds in both intervals are equal. Thus, option (2) is correct.

At $t = 10 \text{ s}$, the velocity changes its sign from positive to negative, which means the particle reverses its direction of motion. Thus, option (3) is incorrect.

The total displacement for the $30 \text{ s}$ interval is the net area under the $v-t$ graph:
Displacement $= \dfrac{1}{2} \times 10 \times 10 - \dfrac{1}{2} \times (30 - 10) \times 20 = 50 - 200 = -150 \text{ m} \neq 0$.
Thus, option (4) is incorrect.