1. **Stating the problem:** We are given a velocity-time graph for a cycle showing velocity changes over 20 seconds. We need to find: a) The total distance travelled. b) The acceleration in the last 8 seconds. 2. **Analyzing the graph:** The velocity graph consists of three parts: - From $t=0$ to $t=4$ seconds, velocity increases linearly from 0 to 10 m/s. - From $t=4$ to $t=12$ seconds, velocity is constant at 10 m/s. - From $t=12$ to $t=20$ seconds, velocity decreases linearly from 10 m/s to 0. 3. **a) Total distance travelled:** Distance is the area under the velocity-time graph. - First part (triangle): $$\text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 10 = 20 \text{ m}.$$ - Second part (rectangle): $$\text{Area}_2 = \text{base} \times \text{height} = (12-4) \times 10 = 8 \times 10 = 80 \text{ m}.$$ - Third part (triangle): $$\text{Area}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (20-12) \times 10 = \frac{1}{2} \times 8 \times 10 = 40 \text{ m}.$$ - Total distance travelled: $$20 + 80 + 40 = 140 \text{ meters}.$$ 4. **b) Acceleration in the last 8 seconds:** The last 8 seconds correspond to time from 12 seconds to 20 seconds. Acceleration is the rate of change of velocity: $$a = \frac{\Delta v}{\Delta t} = \frac{0 - 10}{20 - 12} = \frac{-10}{8} = -1.25 \text{ m/s}^2.$$ Negative acceleration indicates deceleration. **Final answers:** - Total distance travelled: $140$ meters - Acceleration in last 8 seconds: $-1.25$ m/s$^2$