1. **Problem statement:** Given a piecewise linear displacement-time graph, find the velocity and speed for each segment. 2. **Formula:** Velocity is the rate of change of displacement with respect to time, given by $$v = \frac{\Delta d}{\Delta t}$$ Speed is the magnitude of velocity, so $$\text{speed} = |v|$$ 3. **Analyze each segment:** - Segment 1: From $t=0$ to $t=5$ seconds, displacement changes from $0$ m to $5$ m. - Segment 2: From $t=5$ to $t=12$ seconds, displacement remains constant at $5$ m. - Segment 3: From $t=12$ to $t=15$ seconds, displacement changes from $5$ m to $12$ m. - Segment 4: From $t=15$ to $t=20$ seconds, displacement remains constant at $12$ m. - Segment 5: From $t=20$ to $t=30$ seconds, displacement decreases from $12$ m to $0$ m. 4. **Calculate velocity and speed for each segment:** - Segment 1: $$v_1 = \frac{5 - 0}{5 - 0} = \frac{5}{5} = 1\ \text{m/s}$$ Speed $= |1| = 1$ m/s - Segment 2: $$v_2 = \frac{5 - 5}{12 - 5} = \frac{0}{7} = 0\ \text{m/s}$$ Speed $= |0| = 0$ m/s - Segment 3: $$v_3 = \frac{12 - 5}{15 - 12} = \frac{7}{3} \approx 2.33\ \text{m/s}$$ Speed $= |2.33| = 2.33$ m/s - Segment 4: $$v_4 = \frac{12 - 12}{20 - 15} = \frac{0}{5} = 0\ \text{m/s}$$ Speed $= |0| = 0$ m/s - Segment 5: $$v_5 = \frac{0 - 12}{30 - 20} = \frac{-12}{10} = -1.2\ \text{m/s}$$ Speed $= |-1.2| = 1.2$ m/s 5. **Interpretation:** Velocity indicates direction and magnitude of displacement change; speed is the magnitude only. **Final answers:** - Segment 1: velocity = 1 m/s, speed = 1 m/s - Segment 2: velocity = 0 m/s, speed = 0 m/s - Segment 3: velocity = 2.33 m/s, speed = 2.33 m/s - Segment 4: velocity = 0 m/s, speed = 0 m/s - Segment 5: velocity = -1.2 m/s, speed = 1.2 m/s