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A Mathematical Model of Motion
Chapter 5
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
Position x(m) vs Time t(s)
0
20
40
60
1 2 3 4 5 6 7
t(s)
x(m
) Positionx(m)
Describing Motion
1020304050
60d(m)
1 2 3 4 5 6t(s)
A
BC
D
Uniform Motion
Uniform Motion means that equal changes occur during successive time intervals.
Slope
1020304050
60d(m)
1 2 3 4 5 6t(s)
rise Δy
runΔx
slope = rise run
slope = Δy Δx
Slope of Distance vs Time GraphVelocity
slope = Δy Δx
v = Δd Δt
v = d1 – d0
t1 – t0
assume: t0 = 0s
v = d1 – d0
t1 – t0
v = d1 – d0
t1
d1 = d0 + v t1
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
Physics 1-8Practice Problems:1-12
Pages:85, 87, 89Section Review
Page: 89Due: 9/24/02
Problem 12West East
d0= 200v = -15m/s
d0= -400d =d0 + vtv = 12m/s
d = 200 + -15td = -400 + 12t
dtruck = dcar
-400 + 12t =200 + -15t
27t =600
t =22s
d = 200m + (-15m/s)(22s)
d = 130m
0
5
10
15
20
25
30
0 1 2 3 4 5 6 7 8 9 10
t(s)
d(m
)
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
Velocity vs Time Curve
Constant
Faster
Slower
1020304050
60v(m/s)
1 2 3 4 5 6 t(s)
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
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.
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
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
Physics 1-8Practice Probs:13-26
Pages:93,97,98Section Review
Page: 93Due: 9/26/02
v
t
v0d = v0t
d =1/2(v- v0)t
Finding d from V vs t curve
d =1/2(v- v0)t + v0t
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
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
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
d = d0 +1/2(v+v0)t
v2 = v02 +2a(d-d0)
v = v0 + at
d = d0 +v0t + ½at2
*Basic Equations*
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:
v2 = (16m/s)2 +2(4m/s2)(50m)
v2 = v02 +2a(d-d0)
v = 25.6m/s
0
v = √656m2/s2
Physics 3-3
Page:112Problems: 52,54,57
Due: 10/3/06
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
Physics 1-10Practice Probs:27-30
Pages:103Section Review
Page: 103Due: 9/27/02
Falling
Acceleration due toGravity
9.8m/s²32ft/s²
a=g
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
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
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
Physics 3-4Pages:112
Problems:66,67,70Due: 10/10/06
Going straight Up and Down
•Slows down going up.
•Speeds up going down.
•Stops at the top.
•Acceleration is constant.
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 =
Eq: v2 = v02 +2a(d-d0)
0 = v02 +2a(d)
v02 = -2a(d)
= d v0
2 -2a
= d (20m/s)2 -2(-9.8m/s2)
d = 20.4m
v = v0 + at0 = v0 + atv0 = -at
v0 -a
= t
20m/s-(-9.8m/s2)
= t
2.04s = tThe trip up! 4.08s = t
= d (20m/s)2 -2(-9.8m/s2)
d = 20.4m
Physics 1-12Ques: 3-5 Pages:107-8Ques: 15-19 Page:108
Due: 10/2/02
Physics 1-13Ques: 6-11 Pages:10 Ques:39-43 Page:111
Due: 10/3/02Labs Report:10/3/02
Physics 1-14
Ques:44-65 Page:111-114Due: 10/7/02Test: 10/8/02