avg vel

Chapter 2 problem #9b: A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate the average velocity, in m/s. Part (a) of this question was to calculate the average speed which I calculated to be 4.3 m/s, to calculate avg. velocity we use 'dx/'dt = displacement/time elapsed. The book gives an answer to part(b) of 0 m/s, is this because the runner changes directions? As the runner runs around one side of the track and then turns and runs around the other side the directions cancel each other - is this reasoning correct? If not why would the avg. velocity equal 0?

The runner ends up where she started. So her displacement is `ds = 0, and therefore vAve = `ds / `dt = 0..
By contrast her distance is 2 miles, or about 3200 meters. She travels that distance in 12.5 min, or 750 sec, so her average speed is 3200 m / (750 sec) = 4.3 m/s, as you calculated.

avg vel

The runner ends up where she started. So her displacement is `ds = 0, and therefore vAve = `ds / `dt = 0.. By contrast her distance is 2 miles, or about 3200 meters. She travels that distance in 12.5 min, or 750 sec, so her average speed is 3200 m / (750 sec) = 4.3 m/s, as you calculated.

The runner ends up where she started. So her displacement is `ds = 0, and therefore vAve = `ds / `dt = 0..
By contrast her distance is 2 miles, or about 3200 meters. She travels that distance in 12.5 min, or 750 sec, so her average speed is 3200 m / (750 sec) = 4.3 m/s, as you calculated.