1. **Problem statement:** Given a displacement-time graph with piecewise linear segments, find the velocity and speed for each segment. 2. **Key formulas:** - Velocity $v = \frac{\Delta d}{\Delta t}$ (displacement change over time change) - Speed is the magnitude of velocity, so $\text{speed} = |v|$ 3. **Analyze each segment:** - Segment 1: from $t=0$ to $t=5$, displacement changes from $0$ to $5$ meters. $$v_1 = \frac{5 - 0}{5 - 0} = \frac{5}{5} = 1\ \text{m/s}$$ Speed $= 1$ m/s - Segment 2: from $t=5$ to $t=12$, displacement remains constant at $5$ meters. $$v_2 = \frac{5 - 5}{12 - 5} = 0$$ Speed $= 0$ m/s - Segment 3: from $t=12$ to $t=17$, displacement increases from $5$ to $13$ meters. $$v_3 = \frac{13 - 5}{17 - 12} = \frac{8}{5} = 1.6\ \text{m/s}$$ Speed $= 1.6$ m/s - Segment 4: from $t=17$ to $t=20$, displacement remains constant at $13$ meters. $$v_4 = \frac{13 - 13}{20 - 17} = 0$$ Speed $= 0$ m/s - Segment 5: from $t=20$ to $t=30$, displacement decreases from $13$ to $0$ meters. $$v_5 = \frac{0 - 13}{30 - 20} = \frac{-13}{10} = -1.3\ \text{m/s}$$ Speed $= |-1.3| = 1.3$ m/s 4. **Summary:** - Velocities: $v_1=1$, $v_2=0$, $v_3=1.6$, $v_4=0$, $v_5=-1.3$ m/s - Speeds: $1$, $0$, $1.6$, $0$, $1.3$ m/s Velocity indicates direction; speed is always positive.