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.
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You're on a roll — Here's the solution to question 8: The figure shows a velocity-time graph for a car. From the graph, we can identify the following: • Maximum velocity (V_m) = 8 m/s. • Time at which acceleration ends and constant velocity begins = 2 s. • Time at which constant velocity ends and deceleration begins = 6 s. • Total time for the journey = 8 s. a) Find the total distance travelled by the car. The total distance travelled is the area under the velocity-time graph. This graph forms a trapezium. The area of a trapezium is given by the formula: Area = (1)/(2) × (sum of parallel sides) × height In this case: • The length of the top parallel side (duration of constant velocity) = 6 s - 2 s = 4 s. • The length of the bottom parallel side (total time) = 8 s. • The height of the trapezium (maximum velocity) = 8 m/s. Step 1: Calculate the total distance. Total distance = (1)/(2) × (4 s + 8 s) × 8 m/s Total distance = (1)/(2) × 12 s × 8 m/s Total distance = 6 s × 8 m/s Total distance = 48 m The total distance travelled by the car is 48 m. b) Calculate the deceleration of the car. Deceleration occurs in the last phase of the journey, from t=6 s to t=8 s. Deceleration is the negative of the acceleration, which is the gradient of the velocity-time graph during this period. Deceleration = - ( Change in velocityChange in time ) • Initial velocity (u) at t=6 s = 8 m/s. • Final velocity (v) at t=8 s = 0 m/s. • Time interval ( t) = 8 s - 6 s = 2 s. Step 2: Calculate the deceleration. Acceleration = (v - u)/( t) Acceleration = 0 m/s - 8 m/s2 s Acceleration = -8 m/s2 s Acceleration = -4 m/s^2 The deceleration is the magnitude of this negative acceleration. Deceleration = 4 m/s^2 The deceleration of the car is 4 m/s^2. 2 more to go! Send me the next one 📸