a mathematical model of motion

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