motion at constant acceleration - courses.physics.illinois.edu

PHYS 100: Lecture 2

0

t

x vdt∆ = ∫

t

a

a0

t

a0Area = a0t

0

t

v adt∆ = ∫ 0 0v v a t− =

v0

tt

v0Area = v0t

∆v = v-v0

A=(1/2)(∆v)tv0

tt

v0Area = v0t

∆v = v-v0

A=(1/2)(∆v)t

210 0 02

x x v t a t− = +

t

x

dAchilles

Tortoise

t t

x

dAchilles

Tortoise

tReference frame = Earth

t

Tortoise

t t

Tortoise

tReference frame = Achilles

x

Motion at Constant Acceleration

Relative Motion: Reference Frames

Physics 100 Lecture 2, Slide 2

Music

Who is the Artist?

A) Aretha FranklinB) Dinah WashingtonC) Mary WellsD) Etta JamesE) Koko Taylor

BB

Jazz from Etta

Why?She passed last Friday.

The Classic


Rope Around the WorldThe Earth has an equitorial radius of 3963 miles. There are 5280 feet in one mile. Imagine a rope wraped around the equator of a perfectly smooth Earth. Suppose we now add 15 feet to the rope and shape this longer rope into a smoooth circle with its center at the center of the Earth.

How far will the rope now stand away from the surface of the Earth? For now just use your intuition.. Don’t calculate anything, we will do that next.

< 1mm (A) (B) ¼ inch 15 ft (E)2 ft (D)2 inches (C)

BB



How do we start this calculation? Which of the following options would you do first?

Use your calculator to convert radius of Earth to feet. (A)

(B) Use your calculator to determine the initial length of rope in miles

Write down a mathematical expression that relates the radius of the Earth to the original length of rope

(D)

Use your calculator to convert 15 feet into miles (C)

WHY?Always THINK first

Numbers can be substituted in at the end (IF NEEDED)

BB


Rope Around the WorldThe Earth has an equatorial radius of 3963 miles. There are 5280 feet in one mile. Imagine a rope wraped around the equator of a perfectly smooth Earth. Suppose we now add 15 feet to the rope and shape this longer rope into a smoooth circle with its center at the center of the Earth.

Which of the following equations relates RE, the radius of the Earth, to L, the initial length of rope?

L is circumference of circle of radius RE:

(A) none of these(D)2

LR

E= (B)

π

LR

E= (C)

π2

LR

E=

ERL π2=

RE

L

BB



RE

L

Which of the following equations relates ∆L (the extra 15 feet) to ∆R (the amount the new rope stands above the surface of the Earth)?

(A) none of these(D)LL

RR E

∆=∆ (B) (C)LR

LR

E

∆=∆π2

LR

∆=∆

RE

L+∆L

R

∆R = R - RE

LLRRE

∆+=∆+ )(2π

LR ∆=∆π2

ftftL

R 4.22

15

2==

∆=∆

ππ

Look Ma, No Calculator !!

π2

LR

E=

BB


Rope Around the World

< 1mm (A)

The Earth has an equitorial radius of 3963 miles. There are 5280 feet in one mile. Imagine a rope wraped around the equator of a perfectly smooth Earth. Suppose we now add 15 feet to the rope and shape this longer rope into a smoooth circle with its center at the center of the Earth.

How far will the rope now stand away from the surface of the Earth?

(B) ¼ inch 15 ft (E)2 ft (D)2 inches (C)

NOTE: Example of relevance of calculus:

RL π2=π2

1=

dL

dRLR ∆=∆

π2

1

ftft

4.22

15==

π

THIS IS A GREAT EXAMPLE OF “USING MATH AS A TOOL”NOTE THE VALUE OF USING SYMBOLS


1-D Motion Example: Constant Acceleration• An Example of Application of Kinematic Definitions that ARE ALWAYS TRUE !!!

2

021

00tatvxx ++=)(2

00

2

0

2xxavv −+=tavv

00+=

Assume constant acceleration a = a0

t

a

a0

t

a0Area = a0t

0

t

v adt∆ = ∫ 0 0v v a t− =

SPECIAL CASE

0

t

x vdt∆ = ∫

v0

tt

v0Area = v0t

∆v = v-v0

A=(1/2)(∆v)tv0

tt

v0Area = v0t

∆v = v-v0

A=(1/2)(∆v)t

210 0 02

x x v t a t− = +

SPECIAL CASE

dt

xdv

�

�

≡

Velocity:Acceleration:

dt

vda

�

�

≡Everything we did last weekused ONLY these definitions


CheckPoints 1.1 & 1.3

Both True (A) Both False (B)

QUESTION:What are possible problems with these explanations?

• yes because if it is constantly speeding up from rest it should equal out to half the maximum speed half way through the race

• Since the car accelerates at a constant rate, it will reach half of its maximum speed halfway through the race.

A dragster starts from rest and moves with constant acceleration and crosses the finish line (1/4 mile) in 3 seconds..

1. It will reach half of its maximum speed at the 1/8 mile mark .

3. It will reach half of its maximum speed in 1.5 seconds.

1 True (C)3 False

(D) 1 False 3 True

ANSWER:“Halfway Point”: halfway in time is not same as halfway in distance !!

YOU CAN CHECK THIS WITH MOVING MAN !!

Download fromhttp://phet.colorado.edu/en/simulation/moving-man

BB


CheckPoints 1.1 & 1.3: Takeaway Message?BE CAREFUL!

Position, Velocity, Acceleration, Time are different quantities.e.g., “halfway point” NOT well-defined

Halfway in Position

Halfway in Speed

Halfway in Time

Velocity is linear in time

Half speed occurs at half time

Position is NOT linear in time

Half position DOES NOT occur at half time


Follow Up

Both True (A) Both False (B)

A dragster starts from rest and moves with constant acceleration and crosses the finish line (1/4 mile) in 3 seconds..

1. It will reach the 1/8 mile mark when it reaches half of its maximum speed.

3. It will reach the 1/8 mile mark in 1.5 seconds.

1 True (C)3 False

(D) 1 False 3 True

Halfway in Position

Halfway in Speed

Time for Half Position

Time for Half Speed

BB


Relative Motion: Reference Frames

• There is only one equation:ATTA

vvv −=

Check this out at the “Achilles/Tortoise” link in the Lecture column of the Course Planner


CheckPoint 2

xA < 0(A) xA = 0(B) xA > 0(C)

QUESTIONS ABOUT t = 0 in reference frame of B

vA < 0(A) vA = 0(B) vA > 0(C)

xA = -d

vA = +5 m/s

x = 0

BB


CheckPoint 2

QUESTIONS ABOUT t = 0 in reference frame of B

xC < 0(A) xC = 0(B) xC > 0(C)

vC < 0(A) vC = 0(B) vC > 0(C)

xC = +d

vC = -4 m/s

xA = -d

vA = +5 m/sBB


CheckPoint 2

A hits B before C hits B(A)

(B)

(C)

Which ball (A or C) will hit B first?

C hits B before A hits B

C hits B at the same time as A hits B

Let’s draw this from the reference frame of B ! B CA

5 m/s 4 m/s

d d

xC = +dvC = -4 m/s

xA = -dvA = +5 m/s

At t=0: BB


CheckPoint 2

0

10

20

30

40

A B C D E

Which of the graphs to the right represent the motion of balls A and C as observed in the reference frame of ball B?

(e) None of the abovexC = +dvC = -4 m/s

xA = -dvA = +5 m/s

At t=0:

Student reasons:

• Ball C is coming from a positive direction, thus the line should be above the t-axis. Ball A is coming from the negative direction, thus should be below the t-axis. C has a greater velocity than A+B thus it should reach ball B sooner.

• In B's reference frame, C is positive so it starts above the t axis and the opposite is said for A. B "sees" A coming at it with 5m/s and "sees" C moving toward it with 4m/s. So A will have a steeper slope than C

BB


CheckPoint 2

Which of the graphs to the right represent the motion of balls A and C as observed in the reference frame of ball B?

(e) None of the abovexC = +dvC = -4 m/s

xA = -dvA = +5 m/s

At t=0:

speed = magnitude of slope of x vs t plot

velocity = slope of x vs t plot

WHAT DOES MOTION LOOK LIKE IN REFERENCE FRAME OF B?

B CA5 m/s 4 m/s

d d

vA > 0 and vC < 0

Speed(A) > Speed(C)


Notes

• Web Homework 2 DUE TUESDAY 8am

• Office Hours on Monday (consult Website for hours)

• Online Quiz 2 DUE THURSDAY 8pm

• Prelecture 3 and CheckPoint 3 DUE FRIDAY 8am

motion at constant acceleration - courses.physics.illinois.edu

Documents