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
20.a) i) Find the total distance covered by the car in metres. The total distance covered is the area under the velocity-time graph. The graph is a trapezium. Step 1: Identify the shape and its dimensions. The shape is a trapezium with parallel sides corresponding to the total time and the time at constant velocity, and height corresponding to the maximum velocity. The maximum velocity is 60 \, m/s. The total time duration is 30 \, s. The duration of constant velocity is from t=5 \, s to t=25 \, s, which is 25 - 5 = 20 \, s. Step 2: Calculate the area of the trapezium. The formula for the area of a trapezium is A = (1)/(2)(a+b)h, where a and b are the lengths of the parallel sides and h is the perpendicular height. Here, a = 30 \, s, b = 20 \, s, and h = 60 \, m/s. Distance = (1)/(2) (30 \, s + 20 \, s) × 60 \, m/s Distance = (1)/(2) (50 \, s) × 60 \, m/s Distance = 25 \, s × 60 \, m/s Distance = 1500 \, m The total distance covered by the car is 1500 m. ii) Calculate the deceleration of the car. Deceleration is the negative of the acceleration, which is the gradient of the velocity-time graph during the slowing down phase. Step 1: Identify the segment of the graph where deceleration occurs. Deceleration occurs from t=25 \, s to t=30 \, s. Step 2: Determine the initial and final velocities and the time interval for this segment. Initial velocity (u) at t=25 \, s is 60 \, m/s. Final velocity (v) at t=30 \, s is 0 \, m/s. Time interval ( t) is 30 \, s - 25 \, s = 5 \, s. Step 3: Calculate the acceleration. a = (v - u)/( t) a = 0 \, m/s - 60 \, m/s5 \, s a = -60 \, m/s5 \, s a = -12 \, m/s^2 Step 4: State the deceleration. Deceleration is the magnitude of the negative acceleration. Deceleration = |-12 \, m/s^2| = 12 \, m/s^2 The deceleration of the car is 12 m/s^2. b) Determine the time of the day when the matatu caught up with the lorry. Step 1: Define variables and set up equations for the distance covered by the lorry. Let t be the time in hours after 8:00 am when the matatu catches up with the lorry. Lorry's speed (V_L) = 72 \, km/h. Lorry's travel time = t hours. Distance covered by lorry (D_L) = V_L × t = 72t \, km. Step 2: Set up equations for the distance covered by the matatu. Matatu's speed (V_M) = 96 \, km/h. Matatu started at 8:30 am, which is 0.5 hours after the lorry. Matatu's travel time = (t - 0.5) hours. Distance covered by matatu (D_M) = V_M × (t - 0.5) = 96(t - 0.5) \, km. Step 3: Equate the distances and solve for t. When the matatu catches up with the lorry, their distances from Kisumu are equal: D_L = D_M 72t = 96(t - 0.5) 72t = 96t - 48 48 = 96t - 72t 48 = 24t t = (48)/(24) t = 2 \, hours Step 4: Calculate the time of day. The matatu catches up with the lorry 2 hours after 8:00 am. Time of day = 8:00 am + 2 hours = 10:00 am. The matatu caught up with the lorry at 10:00 am. Send me the next one 📸