This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Hey, good to see you again. Step 1: Convert given values to standard units. Total time T = 3 minutes = 3 × 60 s = 180 s. Total distance D = 2.97 km = 2.97 × 1000 m = 2970 m. Maximum speed v_max = 18 m/s. Let a be the acceleration and r be the retardation. The ratio of acceleration to retardation is a:r = 2:3, so r = (3)/(2)a. Step 2: Analyze the motion in three phases using a velocity-time graph. The motion consists of three parts: 1. Uniform acceleration from 0 to v_max. 2. Constant velocity at v_max. 3. Uniform retardation from v_max to 0. Let t_1, t_2, t_3 be the times for acceleration, constant velocity, and retardation, respectively. Let s_1, s_2, s_3 be the distances covered in these phases. For acceleration phase: Initial velocity u=0, final velocity v=v_max=18 m/s. v_max = u + a t_1 18 = 0 + a t_1 t_1 = (18)/(a) The distance covered is the area of the triangle: s_1 = (1)/(2) v_max t_1 = (1)/(2) (18) ((18)/(a)) = (162)/(a) For retardation phase: Initial velocity u=v_max=18 m/s, final velocity $