Genius, specially benchmark graders, plz help?

It is kinda oddly worded, but the answer is B. At any given point in time we can calculate it's average speed so far. Which would be the total distance travled thus far divided by total time thus far. before it rested it has an average speed of 10/3 m/s, but after resting for 2 seconds it changes to 5m/s

The answer is B. You dont find the average of 10, 3, 2, 8, and 4....you must find the 2 different averages of 10, 3 and 8,4....the average of 10 and 3 = 6.5 and the average of 8 and 4 = 6 ( you get the averages by adding all the numbers in the set then dividing by how many there are.) 6.5 is greater then 6...so the average speed is greater before it rested.

The average speed would be total distance divided by total time elapsed.

However, the answers are odd.

a is out.

b is true

c is false

d is also false, as it traveled at 3.33m/sec during the pre-rest, and 2m/sec during the post rest, so the opposite is true, it traveled faster before the rest.

This leaves you with b. You can calculate an average at any point in the bird's course, as I did pre and post rest, but you could also do it at the end of the rest, and you get 2 m/ sec for the flight and rest combined time, or total average of the trip at 2m/sec.

if I'm doing this right the bird traveled a total of 18 meters in 9 seconds. so the average speed would have to be 2 meter per second. but i don't think that has anything to do with it.. the first average speed is 3.33333333 meters per second and after rested is 2 meter per second so i guess you would say changed during trip.... B