text: What you have so far is correct. If the problem is asking for two-way distance (call it s) in terms of F, m and v, then you can substitute F/m for a: \ns = vt = v^2/a = mv^2/F\nNote the numerator is twice the kinetic energy, which checks because work = F*distance. The applied force F did .5mv^2 negative work for distance s/2 and the same magnitude positive work for s/2.\nI hope this is what you were looking for, since otherwise I don't know what your question is.