A particle starts from the origin. Its velocity is given by v(t) = At^2 - Bt , where A and B are positive constants. The particle's turning point is at a distance of 4 units from the origin, and it returns to the origin at t = 3 s. The acceleration of the particle at the instant it returns to the origin is:

Step-by-step solution

Given the velocity function v(t) = At^2 - Bt , the position function x(t) can be found by integrating v(t) with respect to time. x(t) = (At^2 - Bt) dt = A 3 t^3 - B 2 t^2 + C Since the particle starts from the origin, x(0) = 0 , which gives C = 0 . The particle returns to the origin at t = 3 s, so x(3) = 0 : A 3 (27) - B 2 (9) = 0 9A - 4.5B = 0 B = 2A The turning point occurs when the velocity is zero: v(t) = At^2 - 2At = 0 At(t - 2) = 0 Since t > 0 , the turning point is at t = 2 s. The distance of the turning poi