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
You're on a roll — let's tackle these physics problems! Problem 1: Car Motion The problem refers to a graph, but no graph is provided. I will assume the following values based on a common interpretation of such problems and what is faintly visible in the image: The car accelerates from rest to a velocity of 20 m/s in 10 s. It then maintains this constant velocity for 20 s. Finally, it decelerates to rest in 5 s. i) Calculate the total distance traveled by the car: Step 1: Calculate the distance traveled during the acceleration phase (Phase 1). The car accelerates from 0 m/s to 20 m/s in 10 s. This is the area of a triangle on a velocity-time graph. d_1 = (1)/(2) × base × height d_1 = (1)/(2) × 10 s × 20 m/s d_1 = 100 m Step 2: Calculate the distance traveled during the constant velocity phase (Phase 2). The car travels at 20 m/s for 20 s. This is the area of a rectangle. d_2 = velocity × time d_2 = 20 m/s × 20 s d_2 = 400 m Step 3: Calculate the distance traveled during the deceleration phase (Phase 3). The car decelerates from 20 m/s to 0 m/s in 5 s. This is the area of a triangle. d_3 = (1)/(2) × base × height d_3 = (1)/(2) × 5 s × 20 m/s d_3 = 50 m Step 4: Calculate the total distance traveled. D_total = d_1 + d_2 + d_3 D_total = 100 m + 400 m + 50 m D_total = 550 m ii) What is the average speed of the car? Step 1: Calculate the total time taken. T_total = t_1 + t_2 + t_3 T_total = 10 s + 20 s + 5 s T_total = 35 s Step 2: Calculate the average speed. Average speed = Total distanceTotal time Average speed = 550 m35 s Average speed ≈ 15.71 m/s --- Problem 2: Stone Dropped from a Cliff Given: Initial velocity (u) = 0 m/s (since it's dropped) Height of the cliff (s) = 80 m Acceleration due to gravity (g) = 9.8 m/s^2 (standard value) Step 1: Use the kinematic equation for displacement. Since the stone is dropped, its initial velocity is 0 m/s. We use the equation: s = ut + (1)/(2)gt^2 80 m = (0 m/s)t + (1)/(2)(9.8 m/s^2)t^2 80 m = 4.9 m/s^2 · t^2 Step 2: Solve for t^2. t^2 = 80 m4.9 m/s^2 t^2 ≈ 16.3265 s^2 Step 3: Calculate t. t = sqrt(16.3265 s)^2 t ≈ 4.04 s What's next?