Choose the correct statements:

For statement (1): Since the particle moves on a straight line and its average velocity is zero, its net displacement is zero. This implies the particle must have reversed its direction to return to the starting point. For a continuous velocity-time graph (as infinite acceleration is not permitted), the velocity must pass through zero. Thus, statement (1) is incorrect.

For statement (2): If a particle moves in a closed 2D or 3D loop (such as a circle) and returns to its starting point, its net displacement is zero, making the average velocity zero. However, its instantaneous velocity vector can have a non-zero magnitude (speed) throughout the motion. Thus, statement (2) is correct.

For statement (3): \left|\dfrac{d\vec{v}}{dt}\right|$represents the magnitude of total acceleration, while$\dfrac{d}{dt}|\vec{v}|$represents the rate of change of speed (tangential acceleration). In uniform circular motion, the speed is constant, meaning$\dfrac{d}{dt}|\vec{v}| = 0$. However, the direction of velocity changes continuously, resulting in a non-zero centripetal acceleration, meaning \left|\dfrac{d\vec{v}}{dt}\right| \neq 0$. Thus, statement (3) is correct.

For statement (4): The total distance traveled by a particle is always greater than or equal to the magnitude of its net displacement. Dividing both by the same time interval, the average speed is always greater than or equal to the magnitude of the average velocity. Thus, statement (4) is correct.