## College Physics (4th Edition)

The average velocity of the baton is $1.05~m/s$ (north)
We can find the time that each runner spent running. $t_1 = \frac{d_1}{v_1} = \frac{300.0~m}{7.30~m/s} = 41.1~s$ $t_2 = \frac{d_2}{v_2} = \frac{300.0~m}{7.20~m/s} = 41.7~s$ $t_3 = \frac{d_2}{v_2} = \frac{100.0~m}{7.80~m/s} = 12.8~s$ The total time is: $~41.1~s+41.7~s+12.8~s = 95.6~s$ Note that the displacement of the baton from the starting position to the final position is 100 m (north). We can find the baton's average velocity: $v_{av} = \frac{displacement}{time} = \frac{100~m}{95.6~s} = 1.05~m/s$ (north) The average velocity of the baton is $1.05~m/s$ (north)