**Explanation:**

When a particle is executing Simple Harmonic Motion (SHM), its displacement from the mean position follows a sinusoidal function. The maximum displacement of the particle is called the amplitude (A) of the motion. The particle oscillates back and forth between two extreme positions, which are equal in magnitude but opposite in direction.

The time taken for one complete oscillation of the particle is called the time period (T) of the motion. It is the time taken for the particle to go from one extreme position to the other and back again.

The maximum velocity (vmax) of the particle occurs when the particle passes through the mean position (equilibrium position). At this point, the velocity is at a maximum and is in the same direction as the acceleration. The maximum acceleration (amax) of the particle occurs when the particle is at its extreme positions. At these points, the acceleration is at a maximum and is in the opposite direction to the displacement.

If the maximum velocity and acceleration of the particle are equal in magnitude, it means that the amplitude of the motion is equal to the acceleration. Mathematically, this can be represented as:

amax = vmax

Using the equations of SHM, we can relate the amplitude, time period, maximum velocity, and maximum acceleration as follows:

amax = ω^2 A

vmax = ω A

where ω is the angular frequency of the motion.

Since amax = vmax, we can equate the above two equations:

ω^2 A = ω A

Simplifying the equation:

ω^2 = ω

ω = 1

The angular frequency ω is related to the time period T by ω = 2π/T. Substituting this value into the equation:

(2π/T)^2 = 2π/T

Simplifying the equation:

4π^2/T^2 = 2π/T

2π/T = 4π^2/T^2

T = 2

Therefore, the time period of the motion is T = 2 seconds, which is equivalent to 6.28 seconds (approximately). Hence, the correct answer is option A: 6.28 s.