The question asks us to calculate how many kilometers the bus travels between Hawkes Meadow and Brunswick Street, given the average speed and the times for the stops.

Given:

• Average speed of the bus = 29 km/h
• Time at Hawkes Meadow = 14:25
• Time at Brunswick Street = 14:40

Steps to solve:

1. Find the time difference:

   • Time at Brunswick Street: 14:40
   • Time at Hawkes Meadow: 14:25
   • Time difference: 14:40 - 14:25 = 15 minutes

2. Convert time into hours: Since speed is in km/h, we need to convert 15 minutes to hours: $15 \, \text{minutes} = \frac{15}{60} = 0.25 \, \text{hours}$

3. Use the formula for distance: The formula for distance is: $\text{Distance} = \text{Speed} \times \text{Time}$ $\text{Distance} = 29 \, \text{km/h} \times 0.25 \, \text{hours} = 7.25 \, \text{km}$

Final answer:

The bus travels 7.3 km (rounded to 1 decimal place) between Hawkes Meadow and Brunswick Street.

Would you like further details on any part of this?

Here are five related questions for further exploration:

1. How long does it take the bus to travel from Brunswick Street to Upper Parade?
2. What is the total time taken for the bus to travel from Hawkes Meadow to Upper Parade?
3. If the bus's average speed were increased to 35 km/h, how would that affect the travel distance in the same time?
4. What would the travel time be between Hawkes Meadow and Brunswick Street if the bus's speed was 25 km/h?
5. How would a 10-minute delay at one stop affect the overall schedule?

Tip: When solving distance-time-speed problems, always ensure the time units match the speed units!