1. Problem 2: Calculate acceleration, deceleration, and total distance for a train with speed-time graph vertices at (0,0), (20,20), (30,10), (60,0). 2. Formula for acceleration: $$a=\frac{\Delta v}{\Delta t}$$ where $\Delta v$ is change in speed and $\Delta t$ is change in time. 3. Calculate acceleration between 0 and 20 seconds: $$a=\frac{20-0}{20-0}=\frac{20}{20}=1\ \text{m/s}^2$$ 4. Calculate deceleration between 30 and 60 seconds: $$a=\frac{0-10}{60-30}=\frac{-10}{30}=-\frac{1}{3}\ \text{m/s}^2$$ 5. Total distance is area under speed-time graph, sum of trapezoids and triangles: Area 0-20s: triangle $$\frac{1}{2}\times 20 \times 20=200\ \text{m}$$ Area 20-30s: trapezium $$\frac{1}{2}(20+10)\times 10=150\ \text{m}$$ Area 30-60s: triangle $$\frac{1}{2}\times 30 \times 10=150\ \text{m}$$ Total distance $$200+150+150=500\ \text{m}$$ --- 6. Problem 3: Cyclist speed-time graph with segments (0,0) to (20,10) to (40,10) to (60,0). 7. Acceleration 0-20s: $$a=\frac{10-0}{20-0}=\frac{10}{20}=0.5\ \text{m/s}^2$$ 8. Deceleration 20-60s: $$a=\frac{0-10}{60-20}=\frac{-10}{40}=-0.25\ \text{m/s}^2$$ 9. Total distance is area under graph: Area 0-20s: triangle $$\frac{1}{2}\times 20 \times 10=100\ \text{m}$$ Area 20-40s: rectangle $$20 \times 10=200\ \text{m}$$ Area 40-60s: triangle $$\frac{1}{2}\times 20 \times 10=100\ \text{m}$$ Total distance $$100+200+100=400\ \text{m}$$ 10. Average speed over 60s: $$\text{average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{400}{60}=\frac{20}{3}\approx 6.67\ \text{m/s}$$ --- 11. Problem 4: Car speed-time graph with segments (0,5) to (20,25) to (50,25). 12. Acceleration 0-20s: $$a=\frac{25-5}{20-0}=\frac{20}{20}=1\ \text{m/s}^2$$ 13. Distance in first 20s: Area under graph 0-20s is trapezium: $$\frac{1}{2}(5+25)\times 20=300\ \text{m}$$ 14. Total distance in 50s: Area 20-50s is rectangle: $$25 \times 30=750\ \text{m}$$ Total distance $$300+750=1050\ \text{m}$$ 15. Average speed in 50s: $$\frac{1050}{50}=21\ \text{m/s}$$ --- 16. Problem 6: Speed-time graph with segments (0,0) to (20,25) to (40,15) to (60,15). 17. Acceleration 0-20s: $$a=\frac{25-0}{20-0}=\frac{25}{20}=1.25\ \text{m/s}^2$$ 18. Acceleration 20-40s: $$a=\frac{15-25}{40-20}=\frac{-10}{20}=-0.5\ \text{m/s}^2$$ 19. Distance travelled in 60s: Area 0-20s: triangle $$\frac{1}{2}\times 20 \times 25=250\ \text{m}$$ Area 20-40s: trapezium $$\frac{1}{2}(25+15)\times 20=400\ \text{m}$$ Area 40-60s: rectangle $$15 \times 20=300\ \text{m}$$ Total distance $$250+400+300=950\ \text{m}$$ 20. Average speed: $$\frac{950}{60}\approx 15.83\ \text{m/s}$$ --- 21. Problem 8: Car and cyclist speed over 140s. Car speed: rises to plateau then falls to 0 at 140s. Cyclist speed: constant 8 m/s. 22. Car acceleration first 40s: Assuming car speed rises linearly from 0 to plateau speed $v_p$ in 40s. From graph description, plateau speed is unknown, but assume plateau speed $v_p$. Acceleration $$a=\frac{v_p-0}{40}=\frac{v_p}{40}$$ 23. Car deceleration 60-140s: Assuming speed falls from $v_p$ to 0 in 80s. Deceleration $$a=\frac{0-v_p}{140-60}=\frac{-v_p}{80}$$ 24. Cyclist distance in 140s: $$d=8 \times 140=1120\ \text{m}$$ 25. Car distance in 140s: Distance is area under speed-time graph. Assuming trapezoidal shape with plateau from 40 to 60s. Distance $$=\text{area of acceleration} + \text{area of plateau} + \text{area of deceleration}$$ Acceleration area: $$\frac{1}{2} \times 40 \times v_p=20 v_p$$ Plateau area: $$v_p \times (60-40)=20 v_p$$ Deceleration area: $$\frac{1}{2} \times 80 \times v_p=40 v_p$$ Total distance $$20 v_p + 20 v_p + 40 v_p=80 v_p$$ 26. Distance car travelled faster than cyclist: Car speed $>8$ m/s between times $t_1$ and $t_2$. Assuming linear acceleration, solve for $t$ when speed = 8: $$8=\frac{v_p}{40} t \Rightarrow t=\frac{320}{v_p}$$ Similarly for deceleration phase, speed drops from $v_p$ to 0 in 80s. Time when speed = 8: $$8 = v_p - \frac{v_p}{80}(t-60) \Rightarrow t=60 + 80(1 - \frac{8}{v_p})=140 - \frac{640}{v_p}$$ Distance travelled faster than cyclist is area under car speed curve between $t_1$ and $t_2$ minus cyclist speed times same interval. Exact numeric answers require $v_p$ value, which is not given. --- Final answers: Problem 2: Acceleration 0-20s: 1 m/s² Deceleration 30-60s: -1/3 m/s² Total distance: 500 m Problem 3: Acceleration 0-20s: 0.5 m/s² Deceleration 20-60s: -0.25 m/s² Total distance: 400 m Average speed: 6.67 m/s Problem 4: Acceleration 0-20s: 1 m/s² Distance first 20s: 300 m Total distance 50s: 1050 m Average speed 50s: 21 m/s Problem 6: Acceleration 0-20s: 1.25 m/s² Acceleration 20-40s: -0.5 m/s² Distance 60s: 950 m Average speed 60s: 15.83 m/s Problem 8: Cyclist distance 140s: 1120 m Car acceleration first 40s: $\frac{v_p}{40}$ m/s² Car deceleration 60-140s: $-\frac{v_p}{80}$ m/s² Car distance 140s: $80 v_p$ m Distance car faster than cyclist depends on $v_p$ (unknown)