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-v0A=(1/2)(∆v)tv0

tt

v0Area = v0t

∆v = v-v0A=(1/2)(∆v)t

210 0 02x 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

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A)A) StephaneStephane GrappelliGrappelliB)B) Pearl Pearl DjangoDjangoC)C) Mark OMark O’’ConnorConnor’’s Hot Swing Trios Hot Swing TrioD)D) Miles DavisMiles DavisE)E) Cassandra WilsonCassandra Wilson

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

dtxd

vr

r ≡

Velocity:Acceleration:

dtvd

ar

r ≡

202

100 tatvxx ++=)(2 00

2

0

2 xxavv −+=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-v0A=(1/2)(∆v)tv0

tt

v0Area = v0t

∆v = v-v0A=(1/2)(∆v)t

210 0 02x x v t a t− = +

SPECIAL CASE

PreflightsPreflights 1 & 31 & 3

Both True (A) Both False (B)

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A B C D

QUESTION:What are possible problems with these explanations?

• Because the car is accelerating constantly, this means that the velocity of the car is increasing continuously at the same rate, because of this at the halfway point in the race, the car would reach half its minimum speed.

• If the acceleration is constant, its velocity will be continually increasing throughout the race, meaning it will be going relatively "faster" towards the end of the race. This higher velocity means a greater coverage of distance per unit of time, so the racers half speed will occur somewhere before the halfway point in 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

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YOU CAN CHECK THIS WITH MOVING MAN !!

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

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

PreflightsPreflights 1 & 3: Takeaway Message?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 UpFollow 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

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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

Preflight 5Preflight 5

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

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x = 0

Preflight 5Preflight 5

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

Preflight 5Preflight 5

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:

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Preflight 5Preflight 5

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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:

• The answer is C because in both a and b the slope or velocity of ball A is negative and it is moving in a positive direction and the slope or velocity of C is positive but moving in a negative direction. It is not D because C as a steeper slope of 6m/s as opposed to A who has only 5m/s

• A is approaching B at a faster rate (steeper slope) than ball C. This narrows it down to graphs (a) and (d). (d) is then the answer, because A is approaching from a negative disposition.

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Preflight 5Preflight 5

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)