a mathematical model of motion
DESCRIPTION
A Mathematical Model of Motion. Chapter 5. Position Time Graph. d(m). 60. 50. 40. D. 30. B. 20. C. A. 10. 1. 2. 3. 4. 5. 6. t(s). Describing Motion. Uniform Motion. Uniform Motion means that equal changes occur during successive time intervals. d(m). 60. 50. slope = Δy - PowerPoint PPT PresentationTRANSCRIPT
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A Mathematical Model of Motion
Chapter 5
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Position Time Graph
Time t(s) Position x(m)
0.0 10
1.0 12
2.0 18
3.0 26
4.0 36
5.0 43
6.0 48
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Position x(m) vs Time t(s)
0
20
40
60
1 2 3 4 5 6 7
t(s)
x(m
) Positionx(m)
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Describing Motion
1020304050
60d(m)
1 2 3 4 5 6t(s)
A
BC
D
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Uniform Motion
Uniform Motion means that equal changes occur during successive time intervals.
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Slope
1020304050
60d(m)
1 2 3 4 5 6t(s)
rise Δy
runΔx
slope = rise run
slope = Δy Δx
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Slope of Distance vs Time GraphVelocity
slope = Δy Δx
v = Δd Δt
v = d1 – d0
t1 – t0
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assume: t0 = 0s
v = d1 – d0
t1 – t0
v = d1 – d0
t1
d1 = d0 + v t1
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102030405060
d(m)
1 2 3 4 5 6 t(s)
v = d1 – d0
t1 – t0
v = 50m – 20m 5s – 2s
v = 10m/s
d0 = 20m
t1 = 10s
d1 = d0 + v t1d1 = 20m + (10m/s)(10s)
d1 = 120m
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Physics 1-8Practice Problems:1-12
Pages:85, 87, 89Section Review
Page: 89Due: 9/24/02
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Problem 12West East
d0= 200v = -15m/s
d0= -400d =d0 + vtv = 12m/s
d = 200 + -15td = -400 + 12t
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dtruck = dcar
-400 + 12t =200 + -15t
27t =600
t =22s
d = 200m + (-15m/s)(22s)
d = 130m
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0
5
10
15
20
25
30
0 1 2 3 4 5 6 7 8 9 10
t(s)
d(m
)
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Instantaneous Velocity
Distance vs Time
0
1000
2000
3000
4000
1 2 3 4 5 6 7 8 9 10
t(s)
d(m
)
t d
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Velocity vs Time Curve
Constant
Faster
Slower
1020304050
60v(m/s)
1 2 3 4 5 6 t(s)
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1020304050
60v(m/s)
1 2 3 4 5 6t(s)
v = Δd Δt
Δd = vΔt
Area underneath the v vs. t curve is Distance.
A = l x wd = v x t{
v vs t
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Acceleration
Acceleration is the rate of change of velocity.
a = Δv = v1 –v0
Δt t1 – t0
Acceleration is the slope of the velocity vs. time curve.
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Velocity vs. Time
024681012
0 1 2 3 4 5 6 7 8 9 10t(s)
v(m
/s) vΔv=5m/s
Δt=1.5s
Δv=1m/sΔt=8s
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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
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Physics 1-8Practice Probs:13-26
Pages:93,97,98Section Review
Page: 93Due: 9/26/02
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v
t
v0d = v0t
d =1/2(v- v0)t
Finding d from V vs t curve
d =1/2(v- v0)t + v0t
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d =1/2(v-v0)t + v0t
d =1/2(v+v0)t
d = d0 +1/2(v+v0)t
Add Initial Displacement - d0
d =(1/2v)-(1/2v0)t + v0t
d =(1/2v)+(1/2v0)t
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d = d0 +1/2(v + v0)t
v = v0 + at
d = d0 +1/2(v0 + at + v0)t
d = d0 +v0t + ½at2
d = d0 +1/2v0t + 1/2v0t + 1/2at2
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d = d0 +1/2(v+v0)t
Combine: v = v0 + att = (v-v0) /a
d = d0 +1/2(v+v0) (v-v0) /a
v2 = v02 +2a(d-d0)
d = d0 +(v2+v0
2)2a
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d = d0 +1/2(v+v0)t
v2 = v02 +2a(d-d0)
v = v0 + at
d = d0 +v0t + ½at2
*Basic Equations*
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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?
v2 = v02 +2a(d-d0)
V0 = 16m/sa = 4m/s2
d = 50mv = ?
Given:
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v2 = (16m/s)2 +2(4m/s2)(50m)
v2 = v02 +2a(d-d0)
v = 25.6m/s
0
v = √656m2/s2
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Physics 3-3
Page:112Problems: 52,54,57
Due: 10/3/06
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Lab Results
Block Speed vs. Time
0
20
40
60
80
100
120
140
160
180
2000 8
16
24
32
40
48
Time (1/60s)
Sp
ee
d (
cm
/s)
Speed
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Physics 1-10Practice Probs:27-30
Pages:103Section Review
Page: 103Due: 9/27/02
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Falling
Acceleration due toGravity
9.8m/s²32ft/s²
a=g
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t=0s,d=0m,v=0m/st=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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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/s²d0 = 0m v0 = 0m/s d = 55m
Find: t = ?v = ?
Equation: d = d0 + v0t + ½at² d = ½at²
t = √2d/a
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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
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Physics 3-4Pages:112
Problems:66,67,70Due: 10/10/06
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Going straight Up and Down
•Slows down going up.
•Speeds up going down.
•Stops at the top.
•Acceleration is constant.
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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/s²d0 = 0mv = 0m/s
Find: d = ?
t =
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Eq: v2 = v02 +2a(d-d0)
0 = v02 +2a(d)
v02 = -2a(d)
= d v0
2 -2a
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= d (20m/s)2 -2(-9.8m/s2)
d = 20.4m
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v = v0 + at0 = v0 + atv0 = -at
v0 -a
= t
20m/s-(-9.8m/s2)
= t
2.04s = tThe trip up! 4.08s = t
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= d (20m/s)2 -2(-9.8m/s2)
d = 20.4m
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Physics 1-12Ques: 3-5 Pages:107-8Ques: 15-19 Page:108
Due: 10/2/02
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Physics 1-13Ques: 6-11 Pages:10 Ques:39-43 Page:111
Due: 10/3/02Labs Report:10/3/02
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Physics 1-14
Ques:44-65 Page:111-114Due: 10/7/02Test: 10/8/02