Final answer:
Lance Armstrong, Ivan Basso, and Michael Rasmussen's average velocities in the 2005 Tour de France were approximately 41.83 km/h, 41.82 km/h, and 41.80 km/h respectively. These were calculated by dividing the total distance by each rider's respective time, including the time differences between them and Armstrong.
Step-by-step explanation:
To find the average velocity of each rider in the 2005 Tour de France, we will use the formula for velocity, which is the displacement divided by the time taken. The displacement for each rider is the same, which is the total distance of the Tour de France, 3608 km. However, their times will be slightly different due to the time difference in finishing the race. Since average velocity is a vector quantity, it must include direction, but for the purpose of this exercise which only provides a time difference, we'll assume a constant direction and only calculate the magnitude.
Lance Armstrong's time was 86 hours, 15 minutes, and 2 seconds, which we need to convert into seconds or hours for consistency. Ivan Basso's time was 4 minutes and 41 seconds longer than Armstrong's, and Michael Rasmussen's was 11 minutes and 33 seconds longer.
Let's convert Armstrong's time to hours:
86 hours + 15 minutes/60 + 2 seconds/3600 = 86.25056 hours
Lance Armstrong's average velocity = 3608 km / 86.25056 h ≈ 41.83 km/h.
Now let's calculate the times for Basso and Rasmussen:
Ivan Basso's time = Armstrong's time + 4 minutes + 41 seconds
= 86 hours + 15 minutes + 2 seconds + 4 minutes + 41 seconds
= 86.26361 hours
Ivan Basso's average velocity = 3608 km / 86.26361 h ≈ 41.82 km/h.
Michael Rasmussen's time = Armstrong's time + 11 minutes + 33 seconds
= 86 hours + 15 minutes + 2 seconds + 11 minutes + 33 seconds
= 86.29203 hours
Michael Rasmussen's average velocity = 3608 km / 86.29203 h ≈ 41.80 km/h.