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# Sol-Ch03 - CHAPTER 3 E xercises 1 1 0.1 m s 2 a 24 km h 1...

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Exercises 1. 10.1 m / s 2. (a) 24 km / h , 18.7 km / h ; (b) 1.75 h 3. (a) Δx = 320 km; 3.2 h, thus 0.8 h = 48 min. 4. (a) Δx = 720 m + 1920 m = 2640 km, so v =2640 m / 180 s = 14.7 m / s; (b) Δx = 720m – 1920m = –1200 m, so v = –6.67 m / s 5. 38.5 km / 2.817 h = 13.7 km / h or 3.8 m / s 6. (a) 5 m / s ; (b) 2 m / s 7. (a) +5 m / s ; (b) +2.5 m / s ; (c) –5 m / s ; (d) 0 8. Time for Mansell to cover the 315 km distance : T = 315 km / (208 km / h) = 1.514 h = 5.452 × 10 4 s In this time De Cesaris travels 57.53 m/s × T = 313.66 km Thus he loses by 315 km – 313.66 km = 1.34 km 9. B covers 485 km when A finishes, thus its average speed is v B = 485 km / 4 h = 121.25 km / h. Time to complete 500 km is T B = 500 km / v B = 4.124 h = 4 h 7 min 25 s. 10. Distance = 82.67 h × 65.5 km / h = 5415 km. For the Queen Mary v = 5415 km / 92.7 h = 58.4 km / h 11. (a) 6 m / s , (b) 3.5 m / s ; (c) –7 m / s 12. (a) 5 m / s ; (b) 0 ; (c) –10 m / s ; (d) –5 m / s ; (e) 0 13. (a) 550 m / 30 s = 18.3 m / s ; (b) 50 m / 30 s = 1.67 m / s ; (c) –45 m / s / (30 s) = –1.5 m / s 2 14. (a) 2 m / s 2 ; (b) –12.5 m / s 2 ; (c) 200 m / s 2 15. a av = –70 m / s / (0.04 s) = –1750 m / s 2 CHAPTER 3

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16. (a) –3 m / s ; (b) –1.5 m / s 2 17. (a) 0.4 km / 5.4 s = 74.1 m / s ; (b) (410 km / 360 s) / (5.4 s) = 21.1 m / s 2 . 18. Time for total trip should be 4 h. The first leg took 1.8 h = 108 min. Thus second leg must take 240 – 108 – 12 = 120 min. Therefore, v = 120 km / 2 h = 60 km / h 19. (a) 2.83m / s 2 ; (b) 1.88 m / s 2 ; (c) 1.06 m / s 2 ; (d) –6.94 m / s 2 20. (b) +1 m / s ; (c) –5 m / s ; (d) –5 m / s 21. (a) x = 1.04 m to 4.98 m ; (c) 5 m / s ; (d) 5 m / s (b) 0.425 m / 0.1 s = 4.25 m/s (c) 2.25 m / 0.55 s = 5 m/s (d) x (0.5) = 4.3301 m, x (0.501) = 4.3354 m. Thus, Δx /Δt = 0.0052 m / 0.01 s = 5.25 m / s 22. (a) –1.21 m / s ; (b) –1.35 m / s ; (c) –1.34 m / s 23. (a) v = –5 + 6t = 13 m / s, a = 6 m / s 2 ; (b) v = 0 at t = 5 / 6 = 0.83 s 24. (a) (20 m / s) / 5 s = +4 m / s 2 ; (b) 5 m / s 2 ; (c) 2 s 25. (a) 0 ; 3 s ; (b) 3 s ; (c) 2 m / s 2 ; (d) 4 m / s 2 26. (a) 3–4 s ; (b) 0, 7 s ; (c) 1–2 s, 5–6 s ; (d) 4–5 s ; (e) 6–7 s ; (f) 0–1 s; 7–8 s ; (g) 2–3 s 27. (a) 15 m ; (b) 11.7 m / s 28. 27 m , 9 m / s 29. (a) 2.5 m / s ; (b) 5.83 m / s 30. (a) 0; (b) 2.4 m / s 31. (c) 1.67 m / s 2 ; (d) 5 m / s 2
32. (c) 2 m / s 2 ; (d) 4 m / s 2 35. (b) 14.3 mph / s ; (c) 8.4 s ; (d) 13 s at 105 mph 36. (a) 5 s, 40 m ; (b) v A = 8 m / s, v B = 16 m / s 37. (a) 9.2 mph / s, 4.8 mph / s, 3 mph / s; (b) 9.2 mph / s = 13.5 ft / s 2 . Δx = 1/2 (3.5 ft / s 2 ) (17 s) 2 2000 ft; (c) (72 + 240 + 504) mph.s 1200 ft 38. (a) 6.75 × 10 5 m / s 2 ; (b) 1.33 × 10 3 s 39. Δx = v 0 t + 0.5 at 2 : 6 = –2 ×   5 + 12.5 a, so a = 1.28 m / s 2 , E 40. Use Eq. 3.12 with v = 0 and v 0 = 31.1 m / s. (a) 4.11 s , –7.56 m / s 2 ; (b) 6.4 × 10 3 s, –484 m / s 2 41. 96 km/h = 26.7 m / s so a = 5.34 m / s 2 . (a) Δt = (10 m / s) / (5.34 m / s 2 ) = 1.87 s, Δx = (v 2 – v 0 2 ) / 2a = (300 m 2 / s 2 ) / 2 (5.34 m / s 2 ) = 28.1 m; (b) Δt = 1.87 s, Δx = (500 m 2 / s 2 ) / 2a = 46.8 m. 42. (a) 33 s, 5440 m; (b) 3.11 h, 6.37 × 10 6 m; (c) 34.7 d, 4.5 × 10 13 m 43. (a) Using x = 1 / 2 at 2 : t F = 20 s, t R = 24 s, thus Δt = 4 s.

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