First, let's describe the velocity-time graph.

Here's the breakdown of the car's motion and the requested calculations. First, let's describe the velocity-time graph. The graph will have time (s) on the x-axis and velocity (m/s) on the y-axis. Phase 1 (0 to 5 s): The car starts from rest (0 m/s) and accelerates uniformly to 40 m/s. This is represented by a straight line from (0,0) to (5,40). Phase 2 (5 to 15 s): The car travels at a constant velocity of 40 m/s for 10 s. This is represented by a horizontal line from (5,40) to (15,40). Phase 3 (15 to 20 s): The car is brought to rest (0 m/s) in 5 s with uniform retardation. This is represented by a straight line from (15,40) to (20,0). Now, let's determine the requested values: i) The acceleration Step 1: Identify the initial velocity (u), final velocity (v), and time (t) for the acceleration phase (first 5 seconds). u = 0 m/s v = 40 m/s t = 5 s Step 2: Calculate the acceleration using the formula a = (v - u)/(t). a = 40 m/s - 0 m/s5 s a = 40 m/s5 s a = 8 m/s^2 The acceleration is 8 m/s^2. ii) The retardation Step 1: Identify the initial velocity (u), final velocity (v), and time (t) for the retardation phase (last 5 seconds). The car's velocity before retardation is 40 m/s. u = 40 m/s v = 0 m/s t = 5 s Step 2: Calculate the acceleration during this phase. a = (v - u)/(t) a = 0 m/s - 40 m/s5 s a = -40 m/s5 s a = -8 m/s^2 Step 3: State the retardation, which is the magnitude of the negative acceleration. Retardation = |-8 m/s^2| = 8 m/s^2. The retardation is 8 m/s^2. iii) The distance covered after 5 s This is the distance covered during the acceleration phase (first 5 seconds). On a velocity-time graph, distance is the area under the graph. For this phase, the area is a triangle. Step 1: Use the formula for the area of a triangle: Area = (1)/(2) × base × height. Base = 5 s Height = 40 m/s d = (1)/(2) × 5 s × 40 m/s d = (1)/(2) × 200 m d = 100 m The distance covered after 5 s is 100 m. iv) The total distance covered The total distance is the sum of the distances covered in all three phases. Distance in Phase 1 (acceleration, d_1): Calculated above: d_1 = 100 m. Distance in Phase 2 (constant velocity, d_2): This is a rectangle on the graph. Velocity = 40 m/s Time = 10 s d_2 = velocity × time d_2 = 40 m/s × 10 s d_2 = 400 m Distance in Phase 3 (retardation, d_3): This is a triangle on the graph. Base = 5 s Height = 40 m/s d_3 = (1)/(2) × base × height d_3 = (1)/(2) × 5 s × 40 m/s d_3 = 100 m Step 1: Sum the distances from all three phases. Total distance D = d_1 + d_2 + d_3 D = 100 m + 400 m + 100 m D = 600 m The total distance covered is 600 m. That's 2 down. 3 left today — send the next one.