Answer
(a) B
(b) $speed=1~m/s$
(c) $speed=2~m/s$
(d) $speed=0.5~m/s$
Work Step by Step
(a) The slope of a straight-line segment on a position-versus-time graph is equal to the velocity of the object. If the segment points up, the velocity is positive. If the segment points down, the velocity is negative. The speed is the magnitude of the velocity. So, it does not matter if the segment points up or down. We just need to observe the absolute value of the slope of the straight-line segment.
In this case, the player has the greatest speed on B.
(b) The segment A starts with $x_{0}=3~m,~t_{0}=0$ and finishes with $x_{1}=1~m,~t_{1}=2~s$.
$v_{av}=\frac{x_{1}-x_{0}}{t_{1}-t_{0}}=\frac{1~m-3~m}{2~s-0}=-1m/s$
$speed=|-1~m/s|=1~m/s$
(c) The segment B starts with $x_{1}=1~m,~t_{1}=2~s$ and finishes with $x_{2}=3~m,~t_{2}=3~s$.
$v_{av}=\frac{x_{1}-x_{0}}{t_{1}-t_{0}}=\frac{3~m-1~m}{3~s-2~s}=2m/s$
$speed=|2~m/s|=2~m/s$
(d) The segment C starts with $x_{2}=3~m,~t_{2}=3~s$ and finishes with $x_{3}=2~m,~t_{3}=5~s$.
$v_{av}=\frac{x_{3}-x_{2}}{t_{3}-t_{2}}=\frac{2~m-3~m}{5~s-3~s}=-0.5m/s$
$speed=|-0.5~m/s|=0.5~m/s$
