A Mathematical Model of Motion

1. A Mathematical Model of Motion Chapter 5

2. Position Time Graph

4. d(m) 60 50 40 D 30 B 20 C A 10 1 2 3 4 5 6 t(s) Describing Motion

5. Uniform Motion Uniform Motion means that equal changes occur during successive time intervals.

6. d(m) 60 50 slope = Δy Δx slope = rise run rise Δy 40 30 20 run Δx 10 1 2 3 4 5 6 t(s) Slope

7. slope = Δy Δx v = Δd Δt v = d1 – d0 t1 – t0 Slope of Distance vs Time GraphVelocity

8. v = d1 – d0 t1 – t0 v = d1 – d0 t1 d1 = d0 + v t1 assume: t0 = 0s

9. v = 10m/s v = d1 – d0 t1 – t0 d0 = 20m d(m) 60 t1 = 10s 50 40 30 20 10 d1 = d0 + vt1 v = 50m – 20m 5s – 2s 1 2 3 4 5 6 t(s) d1 = 20m + (10m/s)(10s) d1 = 120m

10. Physics 1-8 Practice Problems:1-12 Pages:85, 87, 89 Section Review Page: 89 Due: 9/24/02

11. Problem 12 East West d0= -400 d0= 200 d =d0 + vt v = 12m/s v = -15m/s d = -400+ 12t d = 200+ -15t

12. dtruck = dcar -400+ 12t =200+ -15t 27t =600 t =22s d = 200m+ (-15m/s)(22s) d = 130m

14. Instantaneous Velocity

15. v(m/s) Constant 60 50 Faster 40 Slower 30 20 10 1 2 3 4 5 6 t(s) Velocity vs Time Curve

16. v(m/s) 60 Δd = vΔt 50 v = Δd Δt 40 30 { A = l x w 20 d = v x t 10 1 2 3 4 5 6 t(s) v vs t Area underneath the v vs. t curve is Distance.

17. a = Δv = v1 –v0 Δt t1 – t0 Acceleration Acceleration is the rate of change of velocity. Acceleration is the slope of the velocity vs. time curve.

18. Δv=1m/s Δt=8s Δv=5m/s Δt=1.5s

19. a = Δv Δt a = Δv Δt a = 5m/s 1.5s a = 1m/s 8s a = 3.3m/s² a = 0.13m/s² Find Acceleration from the Graph!! At: t = 1s At: t = 10s

20. Physics 1-8 Practice Probs:13-26 Pages:93,97,98 Section Review Page: 93 Due: 9/26/02

21. Finding d from V vs t curve v d =1/2(v- v0)t d = v0t t v0 d =1/2(v- v0)t + v0t

22. d =1/2(v-v0)t + v0t d =(1/2v)-(1/2v0)t + v0t d =(1/2v)+(1/2v0)t d =1/2(v+v0)t Add Initial Displacement - d0 d = d0 +1/2(v+v0)t

23. v = v0 +at d = d0 +1/2(v + v0)t d = d0 +1/2(v0 +at + v0)t d = d0 +1/2v0t +1/2v0t+ 1/2at2 d = d0 +v0t +½at2

24. v = v0 +at (v2+v02) 2a d = d0 + Combine: t = (v-v0)/a d = d0 +1/2(v+v0)t d = d0 +1/2(v+v0) (v-v0)/a v2 = v02 +2a(d-d0)

25. *Basic Equations* v = v0 + at d = d0 +1/2(v+v0)t d = d0 +v0t +½at2 v2 = v02 +2a(d-d0)

26. A motorcycle traveling at 16 m/s accelerates at a constant rate of 4.0 m/s2 over 50 m. What is its final velocity? Given: V0 = 16m/s a = 4m/s2 d = 50m v = ? v2 = v02 +2a(d-d0)

27. 0 v = √656m2/s2 v2 = v02 +2a(d-d0) v2 = (16m/s)2 +2(4m/s2)(50m) v = 25.6m/s

28. Physics 3-3 Page:112 Problems: 52,54,57 Due: 10/3/06

29. Lab Results

30. Physics 1-10 Practice Probs:27-30 Pages:103 Section Review Page: 103 Due: 9/27/02

32. Falling Acceleration due to Gravity 9.8m/s² 32ft/s² a=g

33. t=0s,d=0m,v=0m/s t=1s,d=4.9m,v=9.80m/s t=2s,d=19.6m,v=19.6m/s t=3s,d=44.1m,v=29.4m/s

35. The Scream Ride at Six Flags falls freely for 31m(62m-205ft). How long does it drop and how fast is it going at the bottom? t = √2d/a Known: a = -g = 9.8m/s² d0 = 0m v0 = 0m/s d = 55m Find: t= ? v = ? Equation: d = d0 + v0t + ½at² d = ½at²

36. t = √2(55m)/9.8m/s² t = 2.51s Equation: v = v0 + at v = at v = (2.51s)(9.8m/s²) v = 24.6m/s = 55mph

37. Physics 3-4 Pages:112 Problems:66,67,70 Due: 10/10/06

38. Going straight Up and Down • Slows down going up. • Stops at the top. • Speeds up going down. • Acceleration is constant.

39. A ball is thrown up at a speed of 20m/s. How high does it go? How long does it take to go up and down? Use up as positive. Known: v0 = 20m/s a = g = -9.8m/s² d0 = 0m v = 0m/s Find: d = ? t =

40. Eq: v2 = v02 +2a(d-d0) v02 -2a = d 0 = v02 +2a(d) v02 = -2a(d)

41. (20m/s)2 -2(-9.8m/s2) = d d = 20.4m

42. v = v0 + at v0 -a 20m/s -(-9.8m/s2) = t = t 0 = v0 + at v0 = -at 2.04s = t 4.08s = t The trip up!

43. (20m/s)2 -2(-9.8m/s2) = d d = 20.4m

44. Physics 1-12 Ques: 3-5 Pages:107-8 Ques: 15-19 Page:108 Due: 10/2/02

45. Physics 1-13 Ques: 6-11 Pages:10 Ques:39-43 Page:111 Due: 10/3/02 Labs Report:10/3/02

46. Physics 1-14 Ques:44-65 Page:111-114 Due: 10/7/02 Test: 10/8/02