In an experiment, a shearwater (a seabird) was taken from its nest, flown a distance 5120 km away, and released. It found its way back to its nest 12.5 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
b. for the whole episode, from leaving the nest to returning

a)   v = -4.74 m / s, b)  v = 0 m / s

Explanation:

Average speed is

          v = Δx /Δt

a) the variation of the distance on the return trip is

          Δx = -5120 km = -5120 10³ km

the negative sign is because the bird is going back

          Δt = 12.5 days (24 h / 1 day) (3600 s / 1 h) = 1.08 10⁶ s

          v = -5120 10³ / 1.08 10⁶

          v = -4.74 m / s

b) the displacement for the round trip is zero, therefore the average velocity is

         v = 0 m / s

hypothesising or postulating

we need to know the time interval during which the force acts. impulse of force = momentum change