text: a physics problem.? a constant f is applied to a mass m intially heading east with v intial until velocity is negative inital the time is 2t what the total distance?      first half of movement.... velocity goes from v to 0 time = t, mass = m, v goes to 0 impulse = ft = m(v - 0) = mat                      (momentum is impluse) a = v/t one way distance = vt - 1/2(v/t)(t^2) = vt - vt/2 = vt/2 two way distance = 2(vt/2) = vt 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:  s = vt = v^2/a = mv^2/f note 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. i hope this is what you were looking for, since otherwise i don't know what your question is.