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
Here are the calculations for the bus journey. Part (a): The maximum speed, in km/h the bus attained. Step 1: Determine the total distance from the area under the speed-time graph. The graph is a trapezium. The area of a trapezium is given by A = (1)/(2)(a+b)h, where a and b are the lengths of the parallel sides and h is the perpendicular height. In this case, the parallel sides are the time intervals where speed is constant (top) and the total time (bottom). The height is the maximum speed (V_max). The duration of constant speed is from t=30 s to t=90 s, so a = 90 - 30 = 60 s. The total time is from t=0 s to t=130 s, so b = 130 - 0 = 130 s. The total distance is given as 2090 m. 2090 = (1)/(2)(60 + 130)V_max 2090 = (1)/(2)(190)V_max 2090 = 95V_max Step 2: Calculate the maximum speed in m/s. V_max = (2090)/(95) V_max = 22 m/s Step 3: Convert the maximum speed from m/s to km/h. To convert m/s to km/h, multiply by (3600)/(1000) or 3.6. V_max = 22 × 3.6 V_max = 79.2 km/h The maximum speed attained is 79.2 km/h. Part (b): The acceleration in the first 30 seconds. Step 1: Identify the initial and final speeds and time for the first 30 seconds. The bus starts from rest, so initial speed u = 0 m/s. The final speed at t=30 s is the maximum speed, v = V_max = 22 m/s. The time taken is t = 30 s. Step 2: Calculate the acceleration. Acceleration is the gradient of the speed-time graph, given by a = change in speedchange in time. a = (v - u)/(t) a = (22 - 0)/(30 - 0) a = (22)/(30) a = (11)/(15) m/s^2 The acceleration in the first 30 seconds is (11)/(15) m/s^2 or approximately 0.733 m/s^2. Part (c): The distance travelled during the last 20 seconds. Step 1: Determine the time interval for the last 20 seconds. The total journey time is 130 seconds. The last 20 seconds are from t = 130 - 20 = 110 s to t = 130 s. Step 2: Calculate the deceleration during the last phase. The deceleration phase is from t=90 s to t=130 s. Initial speed at t=90 s is V_max = 22 m/s. Final speed at t=130 s is 0 m/s. Time taken for deceleration is 130 - 90 = 40 s. Deceleration = (0 - 22)/(40) = -(22)/(40) = -(11)/(20) m/s^2 So, the acceleration is -0.55 m/s^2. Step