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
Right omahendrea@gmail.com, let's go. Step 1: Calculate the acceleration for each car. The formula for constant acceleration starting from rest is a = (v)/(t). For Car A: Initial velocity u_A = 0 m/s Final velocity v_A = 5 m/s Time t_A = 5 s a_A = (v_A)/(t_A) = 5 m/s5 s = 1 m/s^2 For Car B: Initial velocity u_B = 0 m/s Final velocity v_B = 3 m/s Time t_B = 10 s a_B = (v_B)/(t_B) = 3 m/s10 s = 0.3 m/s^2 Step 2: Calculate velocities at five instants of time for plotting. We will calculate velocities at t = 0, 2.5, 5, 7.5, 10 s to cover the relevant time range up to 10 s. The velocity for constant acceleration is v = u + at. Since u=0, v=at. For Car A (a_A = 1 m/s^2): • At t = 0 s, v_A = 1 × 0 = 0 m/s • At t = 2.5 s, v_A = 1 × 2.5 = 2.5 m/s • At t = 5 s, v_A = 1 × 5 = 5 m/s • For t > 5 s, Car A maintains its velocity of 5 m/s. • At t = 7.5 s, v_A = 5 m/s • At t = 10 s, v_A = 5 m/s For Car B (a_B = 0.3 m/s^2): • At t = 0 s, v_B = 0.3 × 0 = 0 m/s • At t = 2.5 s, v_B = 0.3 × 2.5 = 0.75 m/s • At t = 5 s, v_B = 0.3 × 5 = 1.5 m/s • At t = 7.5 s, v_B = 0.3 × 7.5 = 2.25 m/s • At t = 10 s, v_B = 0.3 × 10 = 3 m/s Step 3: Describe the velocity-time graphs. To plot the graphs on the same axes (time on x-axis, velocity on y-axis): • Car A: The graph starts at (0,0). It is a straight line with a positive slope (representing constant acceleration) from (0,0) to (5 s, 5 m/s). After 5 s, the velocity remains constant at 5 m/s, so the graph becomes a horizontal line from (5 s, 5 m/s) to (10 s, 5 m/s). • Car B: The graph starts at (0,0). It is a straight line with a positive slope (representing constant acceleration) from (0,0) to (10 s, 3 m/s). Step 4: Calculate the displacement using the graph. Displacement is the area under the velocity-time graph for the specified time intervals. For Car A (interval 0 s to 5 s): The graph forms a triangle with base 5 s and height 5 m/s. S_A = (1)/(2) × base × height = (1)/(2) × 5 s × 5 m/s = 12.5 m The displacement for Car A is 12.5 m. For Car B (interval 0 s to 10 s): The graph forms a triangle with base 10 s and height 3 m/s.