Q15 SMH i don’t think the intended method was v^2 = n^2..etc, correct me if i’m wrong but the intuitive method is to just let x = Acos(4pi t) solve for t at x =1/4 (t = 1/4pi arcos(1/4A)), then diff for v, set it to half Vmax so u just have -1/2 = sin(arccos(1/4A)) draw up a triangle to get -1/2 = sqrt (16A^2 - 1)/4A, solve for A, then sub it in and derive for acceleration, then the coefficient is just max acceleration.