REPRESENTING MOTION KINEMATICS in One Dimension “To understand motion is to understand nature.” Leonardo da Vinci.

REPRESENTING MOTIONKINEMATICS in One Dimension

“To understand motion is to understand

nature.”Leonardo da Vinci

MECHANICSStudy of motion, force and energy

KinematicsHow

objects move

DynamicsWhy

objects move

Kinematics Objectives

● Represent motion through the use of words, motion diagrams, graphs, and mathematical models.

● Use the terms position, distance, displacement, and time interval in a scientific manner to describe motion.

Motion

• Motion is instinctive–Eyes will notice moving objects more readily than stationary ones

• Object changes position

• Motion can occur in many directions and paths

Representing Motion

• A description of motion relates PLACE and TIME.– Answers the questions WHERE? and WHEN?

PLACE

TIME

• Simplified version of a motion diagram in which the object in motion is replaced by a series of single points

• Size of object must be much less than the distance it moves

Motion Diagram & Particle Model

Describe motion of the car…

• Draw a particle model…

How are the two particle models different? Describe the motion

of each.

A.

B.

Reference FramesAny measurement of position, distance or speed must be made with respect to a frame of reference

80 km/h

Coordinate System

• Tells you the location of the zero point of the variable you are studying and the direction in which the values of the variable increase.

• ORIGIN– The point at which both

variables have the value zero

Distance and Displacement

Distance, d – total ground coveredDisplacement, Dx – change in position of an

object (position is measured from the origin of a chosen coordinate system)

if xxx

Example - A car travels 400 km from Livingston to Philadelphia and then back 200 km to Trenton. What is the displacement of the car? What distance did the car travel?

0 100 200 300 400 kmX =

Livingston PhiladelphiaTrenton

kmkmxxx if 2000200

kmd 600200400

displacement

distance

0

Distance and Displacement

Example - A car travels 400 km from Livingston to Philadelphia and then back 200 km to Trenton. What is the displacement of the car? What distance did the car travel?

0 100 200 300 400 kmX =

Livingston PhiladelphiaTrenton

kmkmxxx if 2000200

kmd 600200400

displacement

distance

0

Distance and Displacement

DisplacementHas magnitude (size) and direction. It is a

VECTOR

1 2 3-1x(m)

mxxx if 213

x

Vectors are represented by

arrows

DisplacementHas magnitude and direction. It is a

VECTOR

1 2 3-1x(m)

mxxx if 231

x

Average Speed and Average Velocity

• Average speed describes how fast a particle is moving. It is calculated by:

• Average velocity describes how fast the displacement is changing with respect to time:

always positivedistanceaverage speed

elapsed time

sign gives direction in 1 Dimension

Scalar (has magnitude only)

Vector (has magnitude and direction)

Average speed and average velocity often have the same magnitude, but not always

t

xv

Example - A car travels 400 km from Philadelphia to Livingston in 2 hours and then back 200 km to Trenton in 1 hr. What is the car’s speed and velocity?

0 100 200 300 400 kmX =

Livingston PhiladelphiaTrenton

hkmt

ds /200

3

600Average

speed

Average velocity

hkmt

xv /67

3

400200

Dx

0

10

20

30

40

50

60

0 2 4 6 8 10

x (k

m)

t (hr)

Runner A

Runner B

ExampleThe position-time graph shows the progress of two runners, A and B.a) When does runner B pass runner A?b) Where does runner B pass runner A? c) What is the starting position for runner A? runner

B?d) After 10 hrs, what is the average velocity of

runner A? After 10 hrs, what is the average velocity of runner B?

e) If the finish line is at 40 km, who won the race?

Analyzing Graphs

• UNITS are the key to analyzing graphs

• When analyzing graphs always check for the following two things:– Slope: Look at the units of the slope to

see if it corresponds to a physically meaningful measurement.

– Area under the curve: look at the units for the area under the curve to see if it corresponds to a measurement.

Graphical Representation of Motion

t

xA

B

Dx

Dt

vt

x

slopeSteepness = speed

Sign = direction

Velocity = speed + direction

Position-Time Graph

slope

Average Velocity from a Graph

t

xA

B

Dx

Dtv

t

xslope

Mathematical Model

0xtvx x = positionx0 = initial positionv = average velocityt = time

v vs. t

0

1

2

3

4

0 1 2 3 4 5 6 7 8 9 10

t (s)

v (m

/s)

x vs. t

0

10

20

30

40

50

0 1 2 3 4 5 6 7 8 9 10

t (s)

x (

m)

Graphs of Motion

Mathematical Model

0xtvx x0 = 20 mv = 2 m/s

Slope

2vMathematical Model

(UNIFORM VELOCITY)

vt

x

tvx AreaArea=20m = Dx

Dx =20m

SL

OP

E

AR

EA

Dx =20m

(init and final positions unknown. ONLY KNOW DISPLACEMENT)

x vs t

-10

-5

0

5

10

15

20

25

0 1 2 3 4 5 6

t (s)

x (

m)

What is happening in this graph?

-10 -5 0 5 10 15 20 mX =

t=0 s STARTEnd t=6 s

v =

1s2s3s4s5s

205 txMathematical

Dia

gra

mm

ati

cG

rap

hic

al

Plot the corresponding v-t graph

0 1 2 3 4 5 6

-10

-5

0

5

10

15

20 x vs. t

t (s)x (m

)

0 1 2 3 4 5 6

-6

-4

-2

0

2v vs. t

t (s)v (m

/s)

x0 = 20 mv = -5 m/s

Slope

smv /5

SL

OP

E

What is happening?

What is happening in each?

A.

B.

C.

D.

0 1 2 3 4 5 6 7 8

-8

-6

-4

-2

0

2

4

6

8

t (s)

v (m

/s)

Draw the corresponding v-t graph

0 1 2 3 4 5 6 7 8

-6

-4

-2

0

2

4

6

8

10 x - t

t (s)x (m

)

v-t S

LO

PE

?v

Which regions shows positive displacement? negative?When is the object moving in the + direction?When is the object moving in the – direction?Which region is the object moving with maximum + velocity?Rank speeds from greatest to leastWhen is the object at rest When does the object change direction?

0 1 2 3 4 5 6 7 8

-6

-4

-2

0

2

4

6

8

10

x - t

t (s)x (m

) CD

E F

AB

A,F B,D,EA,F

B,D,EF

F=D=E>A>B>CC

At 2s, at 7 s

What is the total displacement for the 8 sec? What is the average velocity for the entire 8 sec trip?What is the distance traveled in the 8 sec trip?What is the average speed for the 8 sec trip?

0 1 2 3 4 5 6 7 8

-6

-4

-2

0

2

4

6

8

10

x - t

t (s)x (m

) CD

E F

AB

0m0m/s

32m

4m/s

• GIVEN: s = 6 m/s t = 1 min =60 s

• UNKNOWN: d = ? m

• FORMULA: s = d / t

• SUBSTITUTION: 6 m/s = d / 60 s

• SOLUTION d = 360 m

Problem: A car starting from rest moves with an average speed of 6 m/s. Calculate the distance the car traveled in 1 minute.

• GIVEN: xo = 47m xf = 15m

t = 8 s

• UNKNOWN: Dx = ? m

• FORMULA: Dx=xf-xo

• SUBSTITUTION: Dx= 15-47

• SOLUTION Dx = -32 m

Problem: An object moves from the position +47 m to the position +15 m in 8 s. What is its total displacement? What is its average velocity?

• GIVEN: xo = 47m xf = 15m Dx = -32 m t = 8 s

• UNKNOWN: vav = ? m/s

• FORMULA:• SUBSTITUTION:• SOLUTION

sm

t

xvav

/48

32

0 1 2 3 4 5 6 7 8

-8

-6

-4

-2

0

2

4

6

8

10 x vs. t

t (s)

x (m

)

0 1 2 3 4 5 6 7 8

-8

-6

-4

-2

0

2

4

6

8

t (s)

v (m

/s)

Draw the corresponding x-t graphArea= Dx

-6m

Dx=+12m 3m

Dx=-16m

0 1 2 3 4 5 6 7 8

-8

-6

-4

-2

0

2

4

6

8

10 x vs. t

t (s)

x (m

)

AR

EAv-

t

The Meaning of Shape of a Position-Time graph

Contrast a constant and changing velocity

Contrast a slow and fast moving object

Average Velocity and Instantaneous Velocity

AVERAGE VELOCITY:

-40 -20 0 20 40 60 80 kmX =

-40 -20 0 20 40 60 80 kmX =

Start t=0 End t=2 hr

Average velocity only depends on the initial and final positions. These 2 cars have the same average velocities but different velocities at each instant. When the velocity is not uniform, the instantaneous velocity is not the same as the average velocity.

hrkmxx

t

xv if /60

2

120

2

Acceleration

• Average acceleration describes how quickly or slowly the velocity changes. It is calculated by:

Vector

if

if

tt

vv

t

va

t

va

t

0

lim

• Instantaneous acceleration describes how the velocity changes over a very short time interval:

Acceleration tells us how fast the velocity changes. Velocity tells us how fast the position changes.

SI units: m/s2

Example - A car accelerates along a straight road from rest to 24 m/s in 6.0 s. What is the average acceleration?Average

acceleration

2/4/

46

024sm

s

sm

tt

vv

t

va

if

if

START

v

1s 2s 3s 4s 5s

a

x4 8 12 16

4m/s2

20m/s0

0

START

1s 2s 3s 4s 5s

x

0 1 2 3 4 5

t

x

t=0

t=5

t=4

t=3

t=2t=1

x v aPosition-Time Graph

START

v

1s 2s 3s 4s 5s

a

x

START

v

1s 2s 3s 4s 5s

a

x

Speeding up in + direction

Slowing down in + direction

a and v SAME direction

a and v OPP direction

0

0

START

v

4s 3s 2s 1s 0

a

x

Slowing up in - direction

a and v OPP direction

5s

START

v

4s 3s 2s 1s0

a

x

Speeding up in - direction

a and v SAME direction

5s

• Displacement and velocity are in the direction of motion

• When acceleration is in the SAME direction as velocity, the object is speeding up

• When acceleration is in the OPPOSITE direction to velocity, the object is slowing down

START

1s 2s 3s 4s 5s

x

0 1 2 3 4 5

t

x

t=0

t=5

t=4

t=3

t=2t=1

x v aPosition-Time Graph

Speeding up in + direction

0

START

1s 2s 3s 4s 5s

x

0 1 2 3 4 5

t

+x

t=0

t=5

t=3

t=2

t=1

t=4

x v aPosition-Time Graph

Slowing down in + direction

0

0 1 2 3 4 5t

- x

t=5

t=0t=1t=2

t=3

t=4

x v aSpeeding up in -

direction

START

4s 3s 2s 1s5s

x

0

0 1 2 3 4 5t

- x

START

4s 3s 2s 1s5s

x

t=5

t=0

t=2

t=3t=4

t=1

x v aSlowing down in -

direction

0

0 1 2 3 4 5t

- x

START

4s 3s 2s 1s5s

x

t=5

t=0

t=2

t=3t=4

t=1

x v aSlowing down in -

directionWhat is this object

doing?

0

0 1 2 3 4 5v t

Draw the corresponding

v-t and a-t graphs

x-t graph

v-t graph

0 1 2 3 4 5

x

t

a-t graph

0 1 2 3 4 5t

- x

t=5

t=0t=1t=2

t=3

t=4

x v aSpeeding up in -

direction

START

4s 3s 2s 1s5s

x

0

What is this object doing?

0 1 2 3 4 50

10

20

30 Position vs. Time

t (s)

x (

m)

0 1 2 3 4 5-5

0

5

10 Velocity vs. Time

t (s)

v (

m/s

)

0 1 2 3 4 5-4

-2

0

2Acceleration vs. Time

t (s)

a (

m/s

2)

SLOPE

SLOPE AREA

AREA

PHYSICS DEPARTMENT STORE

-6

-3

0

3

6

9

0 1 2 3 4 5

v (m

/s)

t (s)

v vs. t

0

5

10

15

0 1 2 3 4 5

x (m

)

t (s)

x vs. t

0 1 2 3 4 5

-4

-3

-2

-1

0

1a vs. t

t (s)

a (m

/s2)

AVERAGE VELOCIT

Y

SLOPE =

AREA

RUNNING TOTAL

Dv =-9

vt

x

slo

pe

AVERAGE accelerati

on

SLOPE = at

v

vta

AR

EA

AREA xtv

AREA ]][[ sm No physical meaning

slo

pe

AR

EA

SLOPE = ta /

accelerationConnect with curved line

Constant Acceleration Motion

-6

-3

0

3

6

9

0 1 2 3 4 5

v (m

/s)

t (s)

v vs. t

0

5

10

15

0 1 2 3 4 5

x (m

)

t (s)

x vs. t

0 1 2 3 4 5

-4

-3

-2

-1

0

1a vs. t

t (s)

a (m

/s2)

SLOPE =

AREADv

vt

x

slo

pe

SLOPE = at

v

vtasmssm

]/[]][/[ 2

AR

EA

AREA xtv

slo

pe

AR

EA

vv

constant acceleration aa

Velocity NOT constant

Dx

0

1

2

3

0 1 2 3 4 5

t (s)

v (m

/s)

Estimate the net displacement from 0 s to 5.0 s

Dx = 4.5 + 6 = 10.5 m

Area under v-t curve

0

2

4

6

8

10

0 1 2 3 4 5

x (m

)

t (s)0

2

4

6

8

10

0 1 2 3 4 5

x (m

)

t (s)

0

1

2

3

0 1 2 3 4 5t (s)

v (m

/s)

Construct the corresponding x-t and a-t curves

AR

EA

xtv

at

v

Curved (acceleration)

Straight (constant v)

-1

-0.5

0

0.5

1

0 1 2 3 4 5a (m

/s2 )

t (s)

-1

-0.5

0

0.5

1

0 1 2 3 4 5a (m

/s2 )

t (s)

slo

pe

-2

-1

0

1

2

0 1 2 3 4 t (s)v (m

/s)

Estimate the displacement from 0 s to 4.0 s

Dx = 2 -2 = 0 m

Area under v-t curve

0 1 2 3 4

-3

-2

-1

0

1

2

3

t (s)

a (m

/s2)

0 1 2 3 40

1

2

t (s)

x (m

)

0

1

2

0 1 2 3 4t (s)

x (m

)

AR

EA

-2

-1

0

1

2

0 1 2 3 4t (s)v

(m/s

)

Construct the corresponding x-t and a-t curves

xtv

at

v

-3

-2

-1

0

1

2

3

0 1 2 3 4t (s)

a (m

/s2)

slo

pe

All Curved (acceleration)

Construct the

corresponding x-t and a-t

curves

-6

-4

-2

0

2

4

6

0 1 2 3 4 5 6 7 8 9 10t (s)v

(m/s

)

02468

1012141618

0 1 2 3 4 5 6 7 8 9 10

t (s)

x (m

)

-6

-4

-2

0

2

4

0 1 2 3 4 5 6 7 8 9 10t (s)

a (m

/s2)

curved

curved

curved

straight

straight

at

v

slo

pe

AR

EA

xtv A

B

C

D

E

F

0 1 2 3 4 5v t

x-t graph is quadratic:

v-t graph is linear:

0 1 2 3 4 5

x

t

a-t graph

a = constant

Representations of Accelerated Motion

0vatv

t

Graphical

Mathematical

02 xBtAtx A and B have physical

meaning which we will derive

Mathematical Equations to Represent Constant Accelerated

Motionand relationship to graphs

t

xv

Definition of average velocity:

t

vv

t

vaa f 0

Definition of average acceleration:

tvx

atvv f 0

For constant acceleration:

)( 021

fvvv

constantacceleration

(Slope of x-t graph)

(Slope of v-t graph)

(Equation of v-t graph)

1

2

3

221

0

021

0021

021 )2()()(

attvx

tatvtatvvtvvx f

Mathematical Equations to Represent Constant Accelerated

Motionand relationship to graphs

tvx

atvv f 0

)( 021

fvvv

4

1

KINEMATIC EQUATIONS(Constant Acceleration)

Dx = vDt (definition of average velocity)

v v0 + vf

2=

(average velocity for constant acceleration)

Dv = aDt (definition of avr a)

Dt = (vf – v0)/a

Dx = (v0 + vf)(vf – v0)/2a

5. vf2 = v0

2 + 2aDx(time independent)

KINEMATIC EQUATIONS(Constant Acceleration)

1.

3.

4.

5.

tvx

)( 021

fvvv

atvv f 0

221

0 attvx

xavvf

222

0

2.

0 1 2 3 4 5v t

x-t graph is quadratic:

v-t graph is linear:

0 1 2 3 4 5

x

t

a-t graph

a = constant

Representations of Accelerated Motion

0vatv

t

Graphical

Mathematical

tvatxxattvx

02

21

0

221

0

x = At2 +Bt + x0

x = ½at2 +v0t + x0

Do Now (solve graphically): A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign?

a = -2 m/s2

0

70

0

v (m

/s)

t (s)tstop

AreaDx=36m

Slope=a12

v-t

Example: A car moving at 12 m/s is 36 m away from a stop sign. What acceleration will stop the car exactly at the stop sign?

v0= 12 m/svf = 0a = ? m/s2

Dx = 36 mt

a = -2 m/s2

5-Step Problem Solving

• GIVEN– Read & Draw Diagram (when needed)– List all the given information using variables and units – NO WORDS

• UNKNOWN– What variable are you looking to solve– Variable = ? units

• FORMULA– Write formula

• SUBSTITUTION– Substitute givens into formula (include units)

• SOLUTION – Box your final answer: make sure units are expressed

KINEMATICS PROBLEMS

A passenger jet lands on a runway with a velocity of 70 m/s. Once it touches down, it accelerates at a constant rate of -3 m/s2. How far does the plane travel down the runway before its velocity is decreased to 2 m/s, its taxi speed to the landing gate?vi = 70 m/s

vf = 2a = -3 m/s2

Dx = ? mt

Dx = 816 m

Dx

vi

avf

a

KINEMATICS PROBLEMS

A runner goes 12 m in 3 s at a constant acceleration of 1.5 m/s2. What is her velocity at the end of the 12 m?

vi =vf = ? m/sa = 1.5 m/s2

Dx = 12 mt = 3 s

vi = 1.75 m/svf = 6.25 m/s

The U.S. and South Korean soccer teams are playing in the first round of the world cup. An American kicks the ball giving it an initial velocity of 4 m/s. The ball rolls a distance of 7 m and is then intercepted by a South Korean player. If the ball accelerated at -0.50 m/s2 while rolling across the grass, find its velocity at the time of interception.

KINEMATICS PROBLEMS

vi = 4 m/svf = ? m/sa = -0.50 m/s2

Dx = 7 mt =

vf = 3 m/s