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Ch 2. Motion ina Straight Line

Definitions

1. Kinematics - Motion

Kinetic Energy - Energy associated with motion

2. Motion in physics is broken down into categories

a.) Translational Motion - motion such that an object moves from oneposition to another along a straight line.

b.) Rotational Motion - motion such that an object moves from oneposition to another along a circular path.

c.) Vibrational Motion - motion such that an object moves back andforth in some type of periodicity.

Example: Diatomic Molecule Moving Through Space.

Rotational

Vibrational

Translational

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Dinophysics : Velocity-Raptor

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Speed

1. Speed - How fast an object is moving regardless of what direction it ismoving.

in timeChangeTraveledDistance

=Speed

Example 1. Traveling from your parking space at Conestoga to New YorkCity and back to Conestoga. Find your avg. speed.

x(mi)

y(mi)

up

back

NY

Conestoga

Average Velocity and Displacement

Displacement- Change in position (straight line distance with direction)

Must specify a coordinate system.

2

One way travel = 130 mi.

Total Distance Traveled =260 mi.

Total time elapsed = 5.2 hrs.or (5 hrs 12 min)

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Example: Cartesian coordinate system

x = Change in x = xf

xi

m- 2 m = + 4 m x =6

Avg. Velocity - How fast an object is moving and in what direction it ismoving.

Average Velocity = Change in positionChange in time

Average Velocity =xt

x xt tf i

f i

=

Notation for Displacement & Vel. Example Problem

x = x hat, and has a value of one. The sole purpose of x is toindicate the direction

x(m)

y(m)

xi= 2m

x1

xf= 6m

x2

xi=

xinitial

= Initial Position

xf=xfinal

= Final Position

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Gives the directionDelta

Delta x is the displacement

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Example Problem:

A particle initially at position x = 5 m at time t= 2 s moves toposition x = -2 m and arrives at time t = 4 s.

a.) Find the displacement of the particle.

b.) Find the average speed and velocity of the particle.

x(m)

y(m)

Example Problem 1 revisited

Example 1. Traveling from your parking space at Conestoga to New YorkCity and back to Conestoga. The straight line distance from Conestoga to Yis 97 mi.

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x(mi)

y(mi)

up

back

NY

Conestoga 970

a.) What was the avg. speed from Conestoga to NY?

b.) What was the avg. velocity from Conestoga to NY?

c.) What was the avg. speed for the round trip?

d.) What was the avg. velocity for the round trip?

Note: Speed in PATH DEPENDENT

Velocity is PATH INDEPENDENT. It only depends on the initial and

final positions.

Scalar vs. Vector Quantities

Scalar - Quantity that has magnitude only.

- Mass - Speed

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One way travel = 130 mi.

Total Distance Traveled = 260 mi.

Travel time Con. to NY = 2.6 hrs.Travel time NY to Con. = 2.6 hrs.

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- Length - Energy

Vector - A quantity that has both magnitude and direction.

- Position - Acceleration

- Velocity - Forces

Example: Length vs. Position

x(m)21 3 5 640-2 -1-3-4-5-6

Pt. APt. B

As measured from the origin the length to pt. A is 3m.

To specify the Pt. A in space you must reference it to the origin andthen Pt. A = + (3 m) x

or, the Position VectorA = +(3 m) x

x = x hat, and is called a unit vector in the x-direction. It has a

magnitude of one (hence the name unit) and is used solely to

specify direction.

Position vs Time Graph

x Y1 Y2

Time s Position m Position m

0 0 0

1 1 5

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2 4 10

3 9 15

4 16 20

5 25 25

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Position vs. Time

0

5

10

15

20

25

30

0 1 2 3 4 5 6

Time (sec)

Position(m)(x-coord.)

Movement 1

Movement 2

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A

BC

D

E

F

-150

-100

-50

0

50

100

150

200

250

300

0 10 20 30 40 50 60

Time (s)

Position(

Position vs. Time Graph for a Complete Trip

x yTime (s) Position (m)

0 010 20020 20025 15045 -10060 0

Velocity vs. Time (Constant Velocity)

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Find the average velocity as the objectmoves from:a.) A to B b.) B to C c.) C to D.d.) A to E

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Velocity vs. Time Graph for a Complete Trip

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Velocity Function

0

1

2

3

4

0 1 2 3 4 5

t(sec)

v(m/s)

Position Function

0

1

2

3

4

5

6

7

8

0 1 2 3 4 5

t(sec)

x(m)

x

t

vx

tave =

Area!

x

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A

BC

D

E

F

-150

-100

-50

0

50

100

150

200

250

300

0 10 20 30 40 50 60

Time (s)

Position

(

A B

B C

C

D

DE

E F

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 10 20 30 40 50 60

Time (s)

Velocity(m/s)

Instantaneous Velocity

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Recall:( ) ( )f i

avg

f i

x t x txv x

t t t

= =

vr

(Average velocity)

Consider the function x(t): ( )x t m + 10m

s

t - 0.5m

s

t2

2

4

4=

3

A.avg

50.5 m-35.0 m mv 10.3 x

3.5 s - 2.0 s s= =

r2 s < t < 3.5 s t = 1.5 sec

B.avg

39.7 m-35.0 m mv 23.5 x

2.2 s - 2.0 s s= =

r2 s < t < 2.2 s t = 0.2 sec

0

10

20

30

40

50

60

0 1 2 3 4

t(sec)

x(m)

0

10

20

30

40

50

60

0 1 2 3 4

t(sec)

x(m)

A. B.

t (sec) vave (m/s)

1 17.50.2 23.45

0.01 23.980.001 23.998

Define: The instantaneous velocity at time ti is the slope of the linetangent to the curve X(t) at the time ti.

The instantaneous velocity at the time t = ti is the limiting value we get byletting the upper value of the tf approach ti. Mathematically this isexpressed as:

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slope = 23.5 m/sslope = 10.3m/s

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( )( ) ( ) ( )

v tdX t

dtlim

X t X t

t tt f ti

f i

f i

= =

The velocity function ( )v t is the time derivative of the position function ( )X t

. Differentiation (Calculus)

Acceleration

When the instantaneous velocity of a particle is changing with time, theparticle is accelerating

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f iavg

f i

v vv a x

t tt

= =

vv

(Average Acceleration)

Units:avg

2

m/s mas s

= =v

Example: If a particle is moving with a velocity in the x-direction givenby

t3(t)v 23

=

s

m

a.) What is the average acceleration over the time interval 6 t 12s s

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Example: Instantaneous Acceleration

a.) Find aavg.

over the time interval 5 t 8

3 t 6

b.) What is the acceleration at time t = 6s ?c.) What is the acceleration when the velocity of the particle is zero?

Velocity vs. Time

-15

-10

-5

0

5

10

15

20

25

30

35

0 2 4 6 8 10

time (s)

velocity(m/s)

time (s) vel. (m/s)

0 -10

1 -2

2 -5

3 5

4 12

5 14

6 12

7 218 30

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Positive and Negative Accelerations

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v(m/s)

t (s)

A

C

E

D

B

F

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AB:

BC:

CD:

DE:

EF:

Special Case: Constant Acceleration

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a(m/s2)

t (s)ti= 0 t

f= t

a

0

We make the assumption that the acceleration does not change.Near the surface of the earth, (where most of us spend most ofour time) the acceleration due to gravity is approximately

constant ag

= 9.8 m/s2

Area! Slope!

v(m/s)

t (s)ti= 0 t

f= t

vf

0

x(m)

t (s)ti= 0 t

f= t

xf

xi

Area! Slope!

vi

v = v + a tf i

x x + v tf i i2= +

1

2a t

1.

2.

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Solving for the 3rdconstant acceleration equation

Solve equation 1 for t and substitute t into equation 2 to get the followingequation.

xavv if += 222

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3.

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FREE-FALL ACCELERATION

(9.8 m/s2 = 32 ft/s2)

Consider a ball is thrown straight up.It is in Free Fall the moment it leaves you hand.

Ploty(t) vs. tfor the example above.

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y(t)

tft/2

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Plot v(t) vs. t

FINAL NOTES ON CH 2.

Remember , when going between the following graphs

tf

t/2

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x(t)

AreaUnderCurve

Slope

v(t)

a(t)

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Problem Solving with the constant acceleration equations

1. Write down all three equations in the margin

2. a = 9.8 m/s2 for free fall problems

3. Analyze the problem in terms ofinitialandfinalsections.

CH 2 Describing Motion: Kinematics in 1-D

Practice Questions

_______1. A car starts from Hither, goes 50 km in a straight line to Yon, immediately turnsaround, and returns to Hither. The for this round trip 2 hours. The magnitude of theaverage velocity of the car for this round trip is:

A. 0B. 50 km/hr

C. 100 km/hrD. 200 km/hrE. Cannot be calculated without knowing the acceleration

_______2. An object starts from rest at the origin and moves along thex axis with a constantacceleration of 4 m/s2. Its average velocity as it goes fromx = 2 m tox = 8m is:

A. 1 m/sB. 2 m/sC. 3 m/sD. 5 m/sE. 6 m/s

_______3. A ball is in free fall. Its acceleration is:A. Downward during both ascent and descentB. Downward during ascent and upward during descentC. Upward during ascent and downward during descentD. Upward during both ascent and descentE. Downward at all times except at the very top, when it is zero

_______4. The coordinate-time graph of an object is a straight line with a positive slope. Theobject has:

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x

t

A

x

t

B

x

t

C

x

t

D

x

t

E

A. Constant displacementB. Steadily increasing accelerationC. Steadily decreasing accelerationD. Constant velocityE. Steadily increasing velocity

_______5. A car accelerates from rest on a straight road. A short time later, the car decelerates to astop and then returns to its original position in a similar manner. Which of thefollowing graphs best describes the motion?

Answers: 1.A 2.E 3.A 4.D 5.E

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