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.
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1.1 Describe the motion of the train. The train's motion can be described in five segments: • From t = 0 s to t = 15 s, the train accelerates uniformly from rest to a velocity of 15 m/s. • From t = 15 s to t = 20 s, the train moves at a constant velocity of 15 m/s. • From t = 20 s to t = 28 s, the train accelerates uniformly from 15 m/s to 30 m/s. • From t = 28 s to t = 40 s, the train moves at a constant velocity of 30 m/s. • From t = 40 s to t = 44 s, the train decelerates uniformly from 30 m/s to rest. 1.2 Calculate the acceleration of the train between 20 and 28 s. Step 1: Identify initial and final velocities and times for the interval. At t_i = 20 s, v_i = 15 m/s. At t_f = 28 s, v_f = 30 m/s. Step 2: Use the acceleration formula. a = (v_f - v_i)/(t_f - t_i) a = 30 m/s - 15 m/s28 s - 20 s a = 15 m/s8 s a = 1.875 m/s^2 The acceleration of the train between 20 and 28 s is 1.875 m/s^2. 1.3 Calculate the distance travelled in the 44 s. The distance travelled is the total area under the velocity-time graph. Step 1: Calculate the area for each segment. • Segment 1 (0-15 s, triangle): A_1 = (1)/(2) × base × height = (1)/(2) × 15 s × 15 m/s = 112.5 m • Segment 2 (15-20 s, rectangle): A_2 = width × height = (20 s - 15 s) × 15 m/s = 5 s × 15 m/s = 75 m • Segment 3 (20-28 s, trapezium): A_3 = (1)/(2) × (sum of parallel sides) × height = (1)/(2) × (15 m/s + 30 m/s) × (28 s - 20 s) A_3 = (1)/(2) × 45 m/s × 8 s = 180 m • Segment 4 (28-40 s, rectangle): A_4 = width × height = (40 s - 28 s) × 30 m/s = 12 s × 30 m/s = 360 m • Segment 5 (40-44 s, triangle): A_5 = (1)/(2) × base × height = (1)/(2) × (44 s - 40 s) × 30 m/s = (1)/(2) × 4 s × 30 m/s = 60 m Step 2: Sum all the areas to find the total distance. Total Distance = A_1 + A_2 + A_3 + A_4 + A_5 Total Distance = 112.5 m + 75 m + 180 m + 360 m + 60 m Total Distance = 787.5 m The total distance travelled in 44 s is 787.5 m. 1.4 Draw an acceleration-time graph from this velocity-time graph. To draw the acceleration-time graph, we calculate the acceleration for each segment: • From t = 0 s to t = 15 s: a = 15 m/s - 0 m/s15 s - 0 s = 1 m/s^2. • From t = 15 s to t = 20 s: a = 15 m/s - 15 m/s20 s - 15 s = 0 m/s^2. • From t = 20 s to t = 28 s: a = 30 m/s - 15 m/s28 s - 20 s = 15 m/s8 s = 1.875 m/s^2. • From t = 28 s to t = 40 s: a = 30 m/s - 30 m/s40 s - 28 s = 0 m/s^2. • From t = 40 s to t = 44 s: a = 0 m/s - 30 m/s44 s - 40 s = -30 m/s4 s = -7.5 m/s^2. The acceleration-time graph would consist of horizontal line segments: • A horizontal line at a = 1 m/s^2 from t = 0 s to t = 15 s. • A horizontal line at a = 0 m/s^2 from t = 15 s to t = 20 s. • A horizontal line at a = 1.875 m/s^2 from t = 20 s to t = 28 s. • A horizontal line at a = 0 m/s^2 from t = 28 s to t = 40 s. • A horizontal line at a = -7.5 m/s^2 from t = 40 s to t = 44 s. 2.1 Identify the car with the greatest velocity. Step 1: Understand that velocity on a position-time graph is represented by the gradient (slope) of the line. Step 2: Visually compare the slopes of lines K and L. Line L is steeper than line K. Car L has a greater gradient, indicating a greater velocity. The car with the greatest velocity is Car L. 2.2 How far apart are the two cars at t = 0 s? Step 1: Read the initial position of each car at t = 0 s from the graph. • Car K starts at x = 16 m. • Car L starts at x = 0 m. Step 2: Calculate the difference in their positions. Distance apart = |x_K - x_L| = |16 m - 0 m| = 16 m The two cars are 16 m apart at t = 0 s. 3.1 Draw an accurate displacement-time graph to represent the motion of the car. To draw the graph, plot the given points (Time (s), Displacement (m)) and connect them with straight lines: • (0, 0) • (2, 14) • (4, 28) • (6, 28) • (8, 20) • (10, 12) The graph would show: • A line segment from (0,0) to (2,14). • A line segment from (2,14) to (4,28). • A horizontal line segment from (4,28) to (6,28). • A line segment from (6,28) to (8,20). • A line segment from (8,20) to (10,12). 3.2 Use the graph to determine the velocity of the car after 3 seconds. Step 1: Identify the time interval that includes t = 3 s. This is the interval from t = 2 s to t = 4 s. Step 2: Calculate the gradient (velocity) for this segment using the points (2 s, 14 m) and (4 s, 28 m). v = ( x)/( t) = 28 m - 14 m4 s - 2 s v = 14 m2 s v = 7 m/s The velocity of the car after 3 seconds is 7 m/s. 3.3 What was the car's velocity after 5 seconds? Justify your answer. Step 1: Identify the time interval that includes t = 5 s. This is the interval from t = 4 s to t = 6 s. Step 2: Calculate the gradient (velocity) for this segment using the points (4 s, 28 m) and (6 s, 28 m). v = ( x)/( t) = 28 m - 28 m6 s - 4 s v = 0 m2 s v = 0 m/s The car's velocity after 5 seconds was 0 m/s. This is because the displacement remained constant at 28 m between t = 4 s and t = 6 s, indicating the car was stationary. 3.4 Use the graph to determine the car's velocity after 9 seconds. Step 1: Identify the time interval that includes t = 9 s. This is the interval from t = 8 s to t = 10 s. Step 2: Calculate the gradient (velocity) for this segment using the points (8 s, 20 m) and (10 s, 12 m). v = ( x)/( t) = 12 m - 20 m10 s - 8 s v = -8 m2 s v = -4 m/s The car's velocity after 9 seconds is -4 m/s. 3.5 In words, describe the motion of the car during the 10 seconds. The car starts from rest at the origin and moves in a northerly direction, increasing its speed initially. From t = 0 s to t = 4 s, its displacement increases from 0 m to 28 m. Between t = 4 s and t = 6 s, the car stops and remains stationary at 28 m displacement. From t = 6 s to t = 10 s, the car moves back towards its starting point (in the southerly direction), with its displacement decreasing from 28 m to 12 m. Send me the next one 📸