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
Step 1: Calculate the velocity of the car after accelerating for 30.5 m. The car starts from rest, so its initial velocity (u_1) is 0 m/s. The acceleration (a_1) is 5 m/s^2. The distance traveled (s_1) is 30.5 m. We use the kinematic equation: v_1^2 = u_1^2 + 2a_1s_1. v_1^2 = (0 m/s)^2 + 2(5 m/s^2)(30.5 m) v_1^2 = 10 × 30.5 m^2/s^2 v_1^2 = 305 m^2/s^2 v_1 = sqrt(305) m/s v_1 ≈ 17.464 m/s This velocity v_1 is the initial velocity for the braking phase. Step 2: Calculate the acceleration during braking. For the braking phase, the initial velocity (u_2) is v_1 = sqrt(305) m/s. The car stops, so the final velocity (v_2) is 0 m/s. The time taken to stop (t_2) is 4.0 s. We use the kinematic equation: v_2 = u_2 + a_2t_2. 0 m/s = sqrt(305) m/s + a_2(4.0 s) a_2(4.0 s) = -sqrt(305) m/s a_2 = -sqrt(305)4.0 m/s^2 a_2 ≈ -(17.464)/(4.0) m/s^2 a_2 ≈ -4.366 m/s^2 The acceleration during braking is approximately -4.37 m/s^2. Step 3: Calculate the distance the car travels while braking. We use the kinematic equation: s_2 = (u_2 + v_2)/(2)t_2. s_2 = sqrt(305) m/s + 0 m/s2(4.0 s) s_2 = sqrt(305)2 × 4.0 m s_2 = 2sqrt(305) m s_2 ≈ 2 × 17.464 m s_2 ≈ 34.928 m The distance traveled while braking is approximately 34.9 m. The acceleration while braking is -4.37 m/s^2. The distance traveled while braking is 34.9 m. 3 done, 2 left today. You're making progress.