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To determine the average speed and average velocity of the taxi, we'll use the information provided: 1. **Distance traveled by the taxi (circuitous path):** 23 km 2. **Time taken to travel:** 28 minutes Firstly, let's convert the time from minutes to hours (since speed is typically measured in km/h): \[ \text{Time in hours} = \frac{28 \text{ minutes}}{60} = 0.467 \text{ hours} \] ### Average Speed Average speed is calculated as the total distance traveled divided by the total time taken: \[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \] In this case: \[ \text{Average Speed} = \frac{23 \text{ km}}{0.467 \text{ hours}} \approx 49.26 \text{ km/h} \] ### Average Velocity Average velocity considers both the magnitude and direction of motion. Since the hotel is 10 km away from the station (straight line), and assuming the circuitous path does not significantly alter the direction towards the hotel: \[ \text{Average Velocity} = \frac{\text{Displacement}}{\text{Total Time}} \] Here, displacement is the straight-line distance from the station to the hotel, which is 10 km. \[ \text{Average Velocity} = \frac{10 \text{ km}}{0.467 \text{ hours}} \approx 21.43 \text{ km/h} \] ### Comparison: Average Speed vs. Average Velocity Now, comparing the two: - **Average Speed:** 49.26 km/h - **Average Velocity:** 21.43 km/h The average speed (49.26 km/h) is greater than the average velocity (21.43 km/h). This indicates that while the taxi covered more distance per unit time (due to the circuitous route), the average velocity, which considers the direction of travel towards the hotel, is lower because it only accounts for the displacement towards the destination. Therefore, the two are not equal in this scenario. The average speed reflects the total distance covered over total time, while average velocity considers the displacement over total time, taking direction into account.