In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 km away, and released. The bird found its way back to its nest 13.5 days after release. If we place the origin at the nest and extend the + x-axis to the release point, what was the bird’s average velocity in m/s (a) for the return flight and (b) for the whole episode, from leaving the nest to returning?

Solution 2E Let’s draw a schematic diagram for the motion of the bird. At point a which is 5150 km far from the origin, the bird has released. a) On it’s return flight to it’s nest it must cover a distance of 5150 km. As we know that 1 km = 1000 m. 5150 km = 5150000 meters The time it took to return is 13.5 days. As we know 1 day is 24 hrs, 13.5 days = 13.5 × 24 = 324 hrs = 324 × 3600 s = 1166400 s. the average velocity of the bird is, velocity avg = 5150 km / 324 hrs = 5150000 / 1166400 = 4.415 m/s So, the average velocity of the bird in it’s return trip is 4.415 m/s. b) As, the bird returned to the same place from where she has taken, the total displacement of the bird is zero. And as we know, velocity avg= displacement / time = 0/time = 0. So, the average velocity for the whole trip of the bird is 0 m/s.