Given: Initial velocity u = 0 m/s, acceleration a = 5 m/s2, final velocity v = 25 m/s.

Welcome back A-train✌🏻 — missed you this week. a) To confirm the distance, we calculate the distance covered in three phases: acceleration, constant velocity, and deceleration. Step 1: Calculate distance during acceleration. Given: Initial velocity u = 0 m/s, acceleration a = 5 m/s^2, final velocity v = 25 m/s. Using the kinematic equation v^2 = u^2 + 2as: (25 m/s)^2 = (0 m/s)^2 + 2(5 m/s^2)s_1 625 m^2/s^2 = 10 m/s^2 · s_1 s_1 = (625)/(10) m = 62.5 m Step 2: Calculate distance during constant velocity. Given: Velocity v = 25 m/s, time t = 2100 s. s_2 = v · t s_2 = 25 m/s · 2100 s = 52500 m Step 3: Calculate distance during deceleration. Given: Initial velocity u = 25 m/s, final velocity v = 0 m/s, time t = 25 s. Using the kinematic equation s = ((u+v))/(2)t: s_3 = (25 m/s + 0 m/s)2 · 25 s s_3 = (25)/(2) m/s · 25 s = (625)/(2) m = 312.5 m Step 4: Calculate total distance and convert to kilometers. Total distance S = s_1 + s_2 + s_3: S = 62.5 m + 52500 m + 312.5 m = 52875 m Convert meters to kilometers: S = 52875 m × 1 km1000 m = 52.875 km The distance from the village to the town is indeed 52.875 km. b) Find the cost of fuel needed for a round trip. Step 1: Calculate the total distance for the round trip. Distance one way = 52.875 km. Total distance = 2 × 52.875 km = 105.75 km. Convert to meters: 105.75 km = 105.75 × 1000 m = 105750 m. Step 2: Calculate the total fuel consumed. Fuel consumption rate = 0.2 litres per 500 m. Total fuel = 0.2 L500 m × 105750 m Total fuel = (21150)/(500) L = 42.3 L Step 3: Calculate the total cost of fuel. Cost per litre = 6000 shs. Total cost = 42.3 L × 6000 shs/L = 253800 shs The cost of fuel needed is 253800 shs. c) A hydraulic brake system works based on Pascal's principle, which states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. An illustration would show: • A brake pedal connected to a master cylinder containing brake fluid. • Brake lines extending from the master cylinder to slave cylinders located at each wheel. • Each slave cylinder has pistons that push brake pads against a brake disc (or brake shoes against a drum). When the driver presses the brake pedal, a piston in the master cylinder pushes the brake fluid, creating pressure. This pressure is transmitted equally through the brake lines to the slave cylinders at the wheels. The pistons in the slave cylinders are forced outwards, pressing the brake pads against the rotating brake disc. The resulting friction slows down and stops the wheel. d) The tyre was hot due to the conversion of kinetic energy into heat energy during braking. When the brakes are applied, friction occurs between the brake pads and the brake discs (or shoes and drums). This friction generates a significant amount of heat. This heat is then transferred to the wheel components, including the rims and, subsequently, the tyres through conduction and convection. Additionally, if there was any skidding or hard braking, direct friction between the tyres and the road surface would also generate heat in the tyres. Send me the next one 📸