Recall that n(t) is the derivative of p′/s. This derivative is then divided by s to get curvature. Replace p(t) with p(r(u)), where r is some reparameterization of time. Now p′ is multiplied by r′ by the chain rule, and so is s. The quotient p′/s is the same, but it is a function of u. Differentiate with respect to u and get another factor of r′. Divide by s, and r′ cancels out. Therefore curvature does not depend on the particle's speed.
If the particle moves at constant speed, s = 1, and curvature = |p′′|, the magnitude of the acceleration.
We can use the cross product to develop another formula for curvature. This assumes the path is in 3 space, or can be restricted to 3 space. Remember that acceleration can be split into a tangential and a normal component. We wrote p′′ = n×s + d×s′, where s is speed, d is the unit direction vector, and n is the normal vector. Cross this equation with p′, the velocity vector. Since velocity and direction are parallel, the last term drops out. Since velocity is perpendicular to the normal vector, the norm of that cross product is the norm of p′, times the norm of n, times s. We know the norm of p′ is s, so the norm becomes |n|×s2. Divide by s3 to obtain curvature. Therefore curvature is the norm of (velocity cross acceleration) divided by speed cubed.