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: Describe the velocity-time graph. The velocity-time graph will consist of three segments: • From t=0 s to t=20 s: The train starts from rest (v=0) and accelerates uniformly. This is represented by a straight line segment starting from the origin (0,0) and rising to a maximum velocity, let's call it V, at t=20 s. So, the first point is (0,0) and the second point is (20, V). • From t=20 s to t=50 s: The train travels with uniform velocity V. This is represented by a horizontal straight line segment from (20, V) to (50, V). • From t=50 s to t=60 s: The brakes are applied, and the train decelerates uniformly to rest (v=0) in 10 seconds. This is represented by a straight line segment from (50, V) down to (60, 0). The overall shape of the graph is a trapezoid. Step 2: Calculate the total distance covered. The total distance covered by the train is equal to the area under the velocity-time graph. This area can be divided into three parts: a triangle (acceleration), a rectangle (uniform velocity), and another triangle (deceleration). Area of the first triangle (acceleration phase): A_1 = (1)/(2) × base × height = (1)/(2) × (20 s) × V A_1 = 10V Area of the rectangle (uniform velocity phase): A_2 = base × height = (50 s - 20 s) × V = (30 s) × V A_2 = 30V Area of the second triangle (deceleration phase): A_3 = (1)/(2) × base × height = (1)/(2) × (60 s - 50 s) × V = (1)/(2) × (10 s) × V A_3 = 5V Step 3: Sum the areas to find the total distance. Total distance D = A_1 + A_2 + A_3 D = 10V + 30V + 5V D = 45V Since the maximum velocity V is not given, the total distance is expressed in terms of V. The total distance covered by the train is 45V units. Send me the next one 📸