Ch 2. Motion in a Straight Line

Definitions1. Kinematics - Motion

Kinetic Energy - Energy associated with motion2. Motion in physics is broken down into categories

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

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

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

One Dimensional (x-axis only)

“Dinophysics : Velocity-Raptor”

Straight Line

Spinning

Motion in 3-D space can be complicated

Up and Back

Example: Diatomic Molecule Moving Through Space.

Rotational

Vibrational

Translational

i

f

Net Translation

X - Dir

Note: In this chapter all objects are going to be considered POINT PARTICLES – No Spatial Extent – No Rotations – No Vibrations

Speed

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

Distance TraveledSpeed = Change in time

One way travel = 130 mi.Total Distance Traveled = 260 mi.Total time elapsed = 5.2 hrs. or (5 hrs 12 min)

Example 1 Traveling from your parking space at Conestoga to New York City and back to Conestoga. Find your average speed for the round trip.

x(mi)

y(mi)

up

back

NY

Conestoga 970

≡

Equality by Definition

Speed Calculations are EASY Always distance / time

Round Trip Average Speed

26050

5.2miles

hr

milesspeed

hr

v

Displacement - Change in position (straight line distance with direction) Must specify a coordinate system.

x = Change in x = xf

xi

m- 2 m = + 4 m x 6

∆ “Delta”

Delta x is the displacement or change in the x position

Example: Cartesian coordinate system

up

back

Mathematical Notation for Direction

x(m)

y(m)

xi= 2m

x1

xf= 6m

x2

xi = xinitial= Initial Position

xf = xfinal = Final Position

Average Velocity

Average Velocity = Change in positionChange in time

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

f i

avgf i

x xxv

t t t

≡

Equality by Definition

Notation for Displacement & Velocity = x “hat”, and has a value of one. The sole purpose of is to indicate the

directionExample Problem: A particle initially at position x = 5 m at time t= 2 s moves to position 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.

x̂ x̂

x(m)

y(m)Given:

5 @ 2

2 @ 4i m i s

f m f s

x t

x t

a.)

ˆ ˆ ˆ[ 2 ( 5 )] 7

f i

m m

x x x

x x x x

ix

fx x

ˆ7ˆ3.5

2ms

avg

mavg

s

xv

tx

v x

distanceb.)

t7

3.52

ms

speed

mspeed

s

v

v

Example Problem 1 revisitedExample 1. Traveling from your parking space at Conestoga to New York City

and back to Conestoga. The straight line distance from Conestoga to Y is 97 mi.

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. 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?

Speed in PATH DEPENDENT. Velocity is PATH INDEPENDENT. It only depends on the initial and final positions.

1302.6 50miles

mileshr hr

speedv

ˆ972.6

ˆ37.3mimi

hr hr

xavgv x

2605.2 50miles

mileshr hr

speedv

ˆ05.2 0mi

mihr hr

xavgv

Scalar vs. Vector QuantitiesScalar - Quantity that has magnitude only.

- Mass - Speed- Length - Energy

Vector - A quantity that has both magnitude and direction.- Position - Acceleration- Velocity - Forces

A number (with units) that describes how big or small

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

Pt. APt. B

Example: Length vs. Position

= 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.

x̂

Scalar Vector

ˆ3mA x

ˆ4

4m

m

B x

B

3mA

Concepts Check – The Negatives

Q. Can velocity be negative?

A . YES! Negative displacement means an object moved backwards.

E.g. An object with a displacement ∆x of moved backwards 10m.ˆm -10 x

A. NO! – The least speed an object can have is zero – it is at rest

Q. Can speed be negative?

Q. Can displacement be negative?

A. NO! – The least distance an object can move is zero – it is at rest

Q. Can distance be negative?

A . YES! Negative velocity means an object is moving backwards.

E.g. An object moving is moving backwards with a speed of 10 m/sˆms-10 x

Position vs. Time Graph

Position vs. Time

0

5

10

15

20

25

30

0 1 2 3 4 5 6

Time (sec)

X -

Po

sit

ion

(m

)

x Y1 Y2

Time (s) Position (m)

Position (m)

0 0 0

1 1 5

2 4 10

3 9 15

4 16 20

5 25 25

Both of these movements describe an object moving in one dimension along the x-axis! NOT up and to the right!

x=riset=run

For any time interval

SLOPE is Avg. Velocity

x riseavg t runv slope

0 25m20m15m10m5m

Movement 1

Movement 2

● ● ● ● ●

● ● ● ● ●

1s 2s 3s 4s 5s

1s 2s 3s 4s 5s

Movement 1Movement 2

Position vs. Time Graph for a Complete Trip

A

B C

D

E

F

-150

-100

-50

0

50

100

150

200

250

300

0 10 20 30 40 50 60

Time (s)

Po

sit

ion

(m

)

x y

Time (s) Position (m)

0 0

10 200

20 200

25 150

45 -100

60 0

Find the average velocity as the object moves from:a.) A to B b.) B to Cc.) C to D d.) A to E

200 0

10 0a.) 20m

mss

xA B tv Slope

0 0

20 10b.) 0, Stoppedm

s

xB C tv

150 200

25 20c.) 10m

mss

xD E tv

100 0

45 0d.) 2.22m

mss

xA E tv

Slope of the secant line is vavg

Velocity vs. Time (Constant Velocity)

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

)

t

avg

xv

t

rise Sloperun

Area

x

x

Slope

rise

run

v = height

∆t = base( ) ( )x v t height base area

A

B C

D

E

F

-150

-100

-50

0

50

100

150

200

250

300

0 10 20 30 40 50 60

Time (s)

Po

sit

ion

(m

)

A B

B C

CD

D E

E F

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 10 20 30 40 50 60

Time (s)

Ve

loc

ity

(m

/s)

Velocity vs. Time Graph for a Complete Trip

+X

− X

Velocity vs. Time Graph for a Complete Trip

Area = 200m

Area =

-50m

Area = -250m

∆t = 1.5 sec

0

10

20

30

40

50

60

0 1 2 3 4

t(sec)

x(m

)

●

●●

Instantaneous Velocity recall: f iavg

f i

x t x txˆv x

t t t

(Average velocity)

x t m + 10 m

s t - 0.5

m

s t

22

44

3 avg

50.5 - 35.0ˆv 10.3 x

3.5 - 2.0ms

m m

s s

mavg

39.7 - 35.0ˆv 23.5 x

2.2 - 2.0ms

m

s s

Consider the function x(t): A.

B. ∆t = 0.2 sec

The instantaneous velocity at the time t = ti is the limiting value we get by letting the upper value of the tf approach ti.

Mathematically this is expressed as:

The velocity function is the time derivative of the position function . Differentiation (Calculus)

v t

dX t

dtlim

X t X t

t tt f ti

f i

f i

v t X t

f iavg

f i

v vv ˆa xt tt

avg 2

m/s ma s s

t 3(t)v 23

s

m

6 t 12 s s

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

Units:

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

a.) What is the average acceleration over the time interval

(Average Acceleration)

2 2

2

3243(12) 3(6)(12 6) 6 54

f i

f i

ms

ms

v vvavg t t t

avg s

a

a

Example: Instantaneous Acceleration

a.) Find aavg. over the time interval 5 t 8

b.) What is the acceleration at time t = 6 s ?

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)

ve

loc

ity

(m

/s)

time (s) vel. (m/s)0 -101 -22 -53 54 125 146 127 218 30

2

30 148 5 5.3 m

s

vavg ta

Stopped

Slope of tangent – pick 2 points on the tangent line.Answer will we smaller than the answer to part a, 2(5.3 ).m

s

Objects with zero velocity can be accelerating!0Slope

Positive and Negative Accelerations

v(m/s)

t (s)

A

C

E

D

B

FA→B:

B→C:

C→D:

D→E:

E→F:

Moving Forward

Stopped 0

Moving Backward

Slowing Down, Moving Backward, Pt. B=Stop( ) ( )v negative a slope positive

Slowing Down, Moving Forward, Pt. E=Stop( ) ( )v Positive a slope negative

( ) ( )v Positive a slope positive Speeding Up, Moving Forward

( ) v positive a slope ZERO Constant Speed, Moving Forward

( ) ( )v negative a slope negative Speeding Up, Moving Backward

Special Case: Constant Acceleration

a(m/s2)

t (s)ti = 0 tf = 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 of our time) the acceleration due to gravity is approximately constant ag =

9.8 m/s2

Area! Slope!

v(m/s)

t (s)ti = 0 tf = t

vf

0

x(m)

t (s)ti = 0 tf = t

xf

xi

Area!Slope!

vi

v = v + a tf i

x x + v tf i i2= +

1

2a t

1.

2.

v( )

(

vt

f i f i

a v a t height base

v v a t t

0

)

y b mx

1x2x

1 2

12 ( )i f i

x x x

x v t t v v

12 ( )ix v t t at

xavv if 2223.

Solving for the 3rd constant acceleration equationSolve equation 1 for t and substitute t into equation 2 to get the following equation.

f iv vt

a

212ix v t at

212

f i f iv v v v

i a ax v a

22 21

2 2f i iv v v

f f i ia ax v v v v

2 2 f ia x v v 2 22 2i f f iv v v v 2iv

2 2

2 2

2

2

i f

f i

a x v v

v v a x

tf t/2

y(t)

tf t/2

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.Plot y(t) vs. t for the example above.

Plot v(t) vs. t

2Why? Because 9.8

from the moment it leaves

your hand.

ms

a

iv

20 ftv t

ivGround Level

212yf i iy y v t at

top

Moving up

Stopped 0

Moving down

f iv v at

x(t)

Area Under Curve

Slopev(t)

a(t)

FINAL NOTES ON CH 2.Remember , when going between the following graphs

Problem Solving with the constant acceleration equations1.Write down all three equations in the margin2.a = 9.8 m/s2 for free fall problems3.Analyze the problem in terms of initial and final sections.