In an experiment, a seabird was taken from its nest, flown a distance 5190 km away, and released. It found its way back to its nest a time 13.8 days after release.
What was the bird's average velocity in m/s for the whole episode, from leaving the nest to returning?
Would I double the distance traveled?
Although you could calculate the velocity for the return, there is no information to calculate the velocity for the trip from the nest. How fast did the plane fly?
The time given was the time spent away from the nest after release. We don't know the time spent away from the nest before it was released. I would not double the distance,because the bird did not fly away from the nest; it was
flown.
5190km * 1000m/km = 5,190,000m.
V=(5,190,000m / 13.8days)*(1/24)days /h
* (1/3600)h/s = 4.35m/s.
To find the bird's average velocity in m/s for the whole episode, from leaving the nest to returning, you need to calculate the total displacement (change in position) and divide it by the total time taken.
First, convert the distance traveled from km to meters:
5190 km = 5190 × 1000 = 5,190,000 meters
Next, convert the time taken from days to seconds:
13.8 days × 24 hours/day × 60 minutes/hour × 60 seconds/minute = 1,190,400 seconds
Now you can calculate the average velocity:
Average velocity = Total displacement ÷ Total time taken
Since the bird flew from the nest to the release point and then back to the nest, the total displacement is twice the distance traveled:
Total displacement = 2 × 5,190,000 meters = 10,380,000 meters
Average velocity = 10,380,000 meters ÷ 1,190,400 seconds
Calculating the average velocity:
Average velocity ≈ 8.719 m/s
So, the bird's average velocity for the whole episode, from leaving the nest to returning, is approximately 8.719 m/s.
No, you do not need to double the distance traveled when calculating average velocity. The average velocity is already the rate at which the bird covered the entire distance. It includes both the outbound and return journeys.