Here's the problem: A perfectly flexible cable has length L. Initially, the cable is at rest, with a length x0 of it hanging vertically over the edge of a table. Neglecting friction, compute the length hanging over the edge after a time t. Assume that the sections of the cable remain straight during the motion. The way I see it, the solution will be a formula such that after a certain time t', the solution of the formula is always L, since after t' all of the cable has fallen off of the table. Ignoring a finite L, the physics of the problem should lead to x = |-x0 - .5gt²|, where x is the length of cable hanging over the edge. Setting x equal to L I see that t' = √[2(L-x0)/g]. Therefore I have as a solution the piecewise formula: x = |-x0 - .5gt²| for t < √[2(L-x0)/g] X = L for t ≥ √[2(L-x0)/g] The textbook, on the other hand, has x = x0 Cosh[√(g/L) t]. I don't have the foggiest idea how they ended up with with solution, unless I missed something big and fundamental. A push in the right direction is greatly appreciated.