1. **Problem Statement:** Given a velocity-time graph with three segments: - From $t=0$ to $t=20$ seconds, velocity increases from 25 to 60 m/s. - From $t=20$ to $t=85$ seconds, velocity remains constant at 60 m/s. - From $t=85$ to $t=100$ seconds, velocity decreases from 60 to 0 m/s. Calculate: (i) Acceleration during the first segment. (ii) Retardation during the last segment. (iii) Total distance travelled. (iv) Average speed for the entire journey. --- 2. **Calculate acceleration (i):** Acceleration is the rate of change of velocity. $$a = \frac{\Delta v}{\Delta t} = \frac{60 - 25}{20 - 0} = \frac{35}{20} = 1.75\ \text{m/s}^2$$ 3. **Calculate retardation (ii):** Retardation is negative acceleration during the last segment. $$a = \frac{0 - 60}{100 - 85} = \frac{-60}{15} = -4\ \text{m/s}^2$$ Retardation magnitude is $4\ \text{m/s}^2$. 4. **Calculate total distance travelled (iii):** Distance is the area under the velocity-time graph. - First segment (triangle): $$\text{Area}_1 = \frac{1}{2} \times 20 \times (25 + 60) = 10 \times 85 = 850\ \text{m}$$ - Second segment (rectangle): $$\text{Area}_2 = (85 - 20) \times 60 = 65 \times 60 = 3900\ \text{m}$$ - Third segment (triangle): $$\text{Area}_3 = \frac{1}{2} \times 15 \times 60 = 7.5 \times 60 = 450\ \text{m}$$ Total distance: $$850 + 3900 + 450 = 5200\ \text{m}$$ 5. **Calculate average speed (iv):** Average speed is total distance divided by total time. Total time: $$100 - 0 = 100\ \text{s}$$ Average speed: $$\frac{5200}{100} = 52\ \text{m/s}$$ --- **Final answers:** (i) Acceleration = $1.75\ \text{m/s}^2$ (ii) Retardation = $4\ \text{m/s}^2$ (iii) Total distance travelled = $5200\ \text{m}$ (iv) Average speed = $52\ \text{m/s}$