Suppose that s(t) is like a zigzag with a very small magnitude but very high frequency. Then, s(t) may be as close to zero as you wish, but its derivative is either +(big value), or -(big value), but it is nowhere zero.
Consider the state as a particle. The particle may be at x=0 but it may have a velocity . Now in real sliding mode motion the state is stays within an invariant region around s=0 . In fact it captured in this region where it intersect s=0 with a certain velocity (positive or negative) and the controller function is to ensure that the state is still in this region for all time. The zig-zag motion is an example of a real sliding mode motion where at a certain instant s=0 but ds/dt does not.
The clearest answer is that the sign function is undefined at zero. In practice, the value of the sliding surface will never be zero thanks to effects like quantization and discretization of the controller. Implemented sliding mode control will almost always include a hysteresis around the sign function to prevent infinitely fast switching of the control when the value of the surface is near zero.
Simply the projection of the states through the sliding surface matrix (for linear systems) is zeroed, while the velocity vectors are not zero. To simplify it further, it is similar to the action of the PID controller on the error signal; the integrator action is annihilated when the error signal is zero, but the derivative of the error is not.
from all above should i conclude that surface do not remain uniformly at zero i.e s=0 thats is why we can say analytically that s dot is not equals to zero ???
Simply because sliding occurs on the surface due to a controller of variable structure (usually two) to converge to the surface. Thus, due to switching between structures, the derivative is not zero. It is a type of bang-bang control around a switching manifold.