Physics Acceleration graph
Answer (1)
Answer:Here's the solution to the problem based on the velocity-time graph provided: Assumptions: - The graph is linear between the marked points (A, B, C, D, E, F).- We will estimate the values from the graph as accurately as possible. a) Calculate the total distance covered by the object. The total distance is the sum of the absolute values of the areas under the curve. - Area 1 (0-5s): Triangle above the x-axis. Area = 0.5 * base * height = 0.5 * 5s * 0.8 m/s = 2.0 m- Area 2 (5-10s): Triangle below the x-axis. Area = 0.5 * base * height = 0.5 * 5s * |-0.8 m/s| = 2.0 m- Area 3 (10-15s): Rectangle below the x-axis. Area = base * height = 5s * |-0.4 m/s| = 2.0 m- Area 4 (15-20s): Triangle below the x-axis. Area = 0.5 * base * height = 0.5 * 5s * |-0.4 m/s| = 1.0 m- Area 5 (20-25s): Rectangle above the x-axis. Area = base * height = 5s * 0.8 m/s = 4.0 m Total distance = Area 1 + Area 2 + Area 3 + Area 4 + Area 5 = 2.0 m + 2.0 m + 2.0 m + 1.0 m + 4.0 m = 11.0 m b) Calculate the net displacement covered by the object. The net displacement is the sum of the signed areas under the curve. Areas above the x-axis are positive, and areas below the x-axis are negative. - Area 1 (0-5s): 2.0 m- Area 2 (5-10s): -2.0 m- Area 3 (10-15s): -2.0 m- Area 4 (15-20
