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A Mathematical Model of Motion Chapter 5. Position Time Graph Time t(s)Position x(m) 0.010 1.012 2.018 3.026 4.036 5.043 6.048.

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Presentation on theme: "A Mathematical Model of Motion Chapter 5. Position Time Graph Time t(s)Position x(m) 0.010 1.012 2.018 3.026 4.036 5.043 6.048."— Presentation transcript:

1 A Mathematical Model of Motion Chapter 5

2 Position Time Graph Time t(s)Position x(m) 0.010 1.012 2.018 3.026 4.036 5.043 6.048

3

4 Describing Motion 10 20 30 40 50 60 d(m) 123456 t(s) A B C D

5 Physics 1-8 Practice Problems:1-3 Page:85 Due: 9/25/01

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

7 Slope 10 20 30 40 50 60 d(m) 123456 t(s) rise Δy run Δx slope = rise run slope = Δy Δx

8 Slope of Distance vs Time Graph Velocity slope = Δy Δx v = Δd Δt v = d 1 – d 0 t 1 – t 0

9 assume: t 0 = 0s v = d 1 – d 0 t 1 – t 0 v = d 1 – d 0 t1 d 1 = d 0 + v t 1

10 10 20 30 40 50 60 d(m) 123456 t(s) v = d 1 – d 0 t 1 – t 0 v = 50m – 20m 5s – 2s v = 10m/s d 0 = 20m t 1 = 10s d 1 = d 0 + v t 1 d 1 = 20m + ( 10m/s)(10s) d 1 = 120m

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

12 Problem 12 West East d 0 = 200 v = -15m/s d 0 = -400 d =d 0 + vt v = 12m/s d = 200 + -15td = -400 + 12t

13 d truck = d car -400 + 12t =200 + -15t 27t =600 t =22s d = 200m + (-15m/s)(22s) d = 130m

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15 Instantaneous Velocity

16 Velocity vs Time Curve Constant Faster Slower 10 20 30 40 50 60 v(m/s) 1234 56 t(s)

17 10 20 30 40 50 60 v(m/s) 123456 t(s) v = Δd Δt Δd = vΔt Area underneath the v vs. t curve is Distance. A = l x w d = v x t { v vs t

18 Acceleration Acceleration is the rate of change of velocity. a = Δv = v 1 –v 0 Δt t 1 – t 0 Acceleration is the slope of the velocity vs. time curve.

19 Δv=5m/s Δt=1.5s Δv=1m/s Δt=8s

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

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

22 v t v0v0 d = v 0 t d =1/2(v- v 0 )t Finding d from V vs t curve d =1/2(v- v 0 )t + v 0 t

23 d =1/2(v+v 0 )t d = d 0 +1/2(v+v 0 )t Add Initial Displacement - d 0 d =(1/2v)-(1/2v 0 )t + v 0 t d =(1/2v)+(1/2v 0 )t

24 d = d 0 +1/2(v + v 0 )t v = v 0 + at d = d 0 +1/2(v 0 + at + v 0 )t d = d 0 +v 0 t + ½at 2 d = d 0 +1/2v 0 t + 1/2v 0 t + 1/2at 2

25 d = d 0 +1/2(v+v 0 )t Combine: v = v 0 + at t = (v-v 0 ) /a d = d 0 +1/2(v+v 0 ) (v-v 0 ) /a v 2 = v 0 2 +2a(d-d 0 ) d = d 0 + (v 2 +v 0 2 ) 2a

26 d = d 0 +1/2(v+v 0 )t v 2 = v 0 2 +2a(d-d 0 ) v = v 0 + at d = d 0 +v 0 t + ½at 2 *Basic Equations*

27 A motorcycle traveling at 16 m/s accelerates at a constant rate of 4.0 m/s 2 over 50 m. What is its final velocity? v 2 = v 0 2 +2a(d-d 0 ) V 0 = 16m/s a = 4m/s 2 d = 50m v = ? Given:

28 v 2 = (16m/s) 2 +2(4m/s 2 )(50m) v 2 = v 0 2 +2a(d-d 0 ) v = 25.6m/s 0 v = √ 656m 2 /s 2

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

30 Lab Results

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

32

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

34 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

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36 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? Known: a = -g = 9.8m/s² d 0 = 0m v 0 = 0m/s d = 55m Find: t = ? v = ? Equation: d = d 0 + v 0 t + ½at² d = ½at² t = √ 2d/a

37 t = √ 2(55m)/9.8m/s² t = 2.51s Equation: v = v 0 + at v = at v = (2.51s)(9.8m/s²) v = 24.6m/s = 55mph

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

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

40 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: v 0 = 20m/s a = g = -9.8m/s² d 0 = 0m v = 0m/s Find: d = ? t =

41 Eq: v 2 = v 0 2 +2a(d-d 0 ) 0 = v 0 2 +2a(d) v 0 2 = -2a(d) = d v 0 2 -2a

42 = d (20m/s) 2 -2(-9.8m/s 2 ) d = 20.4m

43 v = v 0 + at 0 = v 0 + at v 0 = -at v 0 -a = t 20m/s -(-9.8m/s 2 ) = t 2.04s = t The trip up! 4.08s = t

44 = d (20m/s) 2 -2(-9.8m/s 2 ) d = 20.4m

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

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

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


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