September 15, 2009Physics I Lesson 2 Dr J. Tison 1 Kinematics: One Dimensional Motion.

September 15, 2009 Physics I Lesson 2Dr J. Tison

1

Kinematics: One Dimensional Motion


2

Short History of Classical Mechanics:Greeks – 17th and 18th Century

• Aristotle and other Greek philosophers– Theorized, no experimental data

• Muslim physicists in 11th and 12th centuries– Scientific methods first applied

• Inertia, momentum, force and mass, gravity

– Precursor to Newton’s Laws of Motion

• Modern interpretation of principles of motion– Keppler (-1630)

• Planetary motion observations by Tyco Brahe

– Galileo (-1642)• Acceleration of rolling ball on inclined plane

– Newton (-1727)• Principia… Laws of Motion


3

Classical Mechanics

• Kinematics– How objects move– Equations relating

• Distance, velocity, acceleration, time

• Dynamics– Why things move– Equations relating

• Momentum, force, potential energy, pressure and power.


4

Motion Types

• Translational– 1 Dimension (this lesson)

• Straight line motion

– 2 & 3 Dimension (later lesson)• Trajectory motion

• Rotational (later lessons)

• Combined Translation and Rotation (later lessons)


5

1D Motion:Model and Concepts

• ‘Idealized’ particle– Mathematical point– No spatial extent (no size)


6

1D Motion:Frame of Reference

• Measurements (position, distance, speed)– Requires reference point– Observer vs Object


7

1D Motion: Frame of Reference

• Coordinate System– Cartesian coordinate system (x,y axes)

• x (horizontal motion); y (vertical motion)

– Position, and distance along x or y axis – Displacement vs distance

• Change in position vs how far object moved


8

Motion along a Line• Distance moved along

x axis– ∆x = x2-x1

• Time moved along x axis– ∆t = t2-t1

• Average speed moved along x axis– ∆v = ∆x/∆t

∆ is symbol for ‘change’:final value less

initial value (displacement)

2010 30 x400

y

Distance (m)

S

E

N

W-20 -10 302010-30

Distance =a+b

Displacement =bb-aa

bb

aa

ba

(x1, t1) (x2, t2)

40


9

Displacement Vector• Vectors vs Scalars

– Vector symbol• V V (B,I,S)(B,I,S)

– Magnitude/direction

• Business woman walking– Displacement vectors

• aa = 30m east• bb = 60m west• Net displacement vector (cc)

a a + bb = 30m west

– Total distance (not vector)• |b|+ |a| = 90m

S

E

N

W-20 -10 30m2010-30

bb

aa

cc


10

Speed & Velocity

• Speed (scalar)– How far object travels in a given time interval

• average speed = (distance traveled)/(time elapsed)

• Velocity (vector)– How far and in what direction traveled in given

time interval• average velocityaverage velocity = (displacementdisplacement)/(time

elapsed)

• Vector = scalar (magnitude) + direction


11

Average Speed & Average Velocity:Walking Businesswoman

• Average speed– ∆x(distance)/∆t = [|x2- x1| + |x3- x2|]/|t3- t1|

= (30+60)/(90)

= 1m/s

• Average velocity– ∆xx(displacement)/∆t = (x3- x1)]/(t3- t1)

= (-20-10)/(90)

= -30/90

= -0.33m/s (West)

S

N

W-20 -10 302010-30

bb

aa

40m

tt1=0 at xx1=10m

tt2=30s at xx2= 40m

tt3=90s at xx3=-20m

xx33

xx11 xx22


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Instantaneous Velocity• Constant or changing

velocity• Instantaneous velocity is..

– Velocity at an instant of time

– “Average velocity during an infinitesimal time interval”

• v =lim(∆t →0)[∆x/∆t]

v: instantaneous

vv: average

lim∆t →0

Vel

oci

ty (

mp

h)V

elo

city

(m

ph)

Time (h)Time (h)

2020

00 0.10.1 0.40.40.20.2 0.30.3

4040

6060

Time (h)Time (h)

2020

00 0.10.1 0.40.40.20.2 0.30.3404

0606

0V

elo

city

(m

ph)

Vel

oci

ty (

mp

h)

Average velocityAverage velocity

●● Instantaneous velocityInstantaneous velocity

lim∆t →0


13

Acceleration

• Changing velocity = acceleration– Average acceleration = (change of velocity/(time elapsed)

– aa = (v2-v1)/(t2 - t2) = ∆v/∆t

• Instantaneous acceleration– a =lim(∆t →0)[∆v/∆t]

tt1 =0, =0, v1 =0 =0

t =1.0S, t =1.0S, v = = 15mph15mph

t =2s, t =2s, v1 =30mph =30mph t =5s, t =5s, v =60mph =60mph

a =


14

Acceleration vs velocity

• Acceleration » speeding up– How quickly velocity changes

• Velocity – How quickly position changes

• Deceleration (a vector) » slowing down – Frame of reference »

• +a to right,

• - a to left

• Acceleration vector is opposite to velocity vector


15

Constant Acceleration

• Equations relating x, v, a, and t– Initial conditions

• (t1, x1) = (t0, x0) = (0,0)

• (t2, x2) = (t, x)

• Condition of motion– Magnitude of acceleration is constant– Motion in a straight line.

» aa (instantaneous) = a (average)


16


vv = ∆xx/∆t =(x x -xx0)/t

» x» x = xx0+vv ttt

xx

xx0

xx

tt

vv

vv0

vv

tt

∆x = x-x0

∆t = t-t0 = t

t =0t =0

t =0t =0

∆v = v-v0

∆t = t-t0 = t

tt

aa = ∆vv/∆t = (v v -vv0)/t

» v» v = vv0+aa t


17


xx = x0+ <v v >t; – where <v v > is average

velocity– From time t = 0 → t = t,

• velocity goes from vv0 → vv

<v v > = (vv0 +vv)/2

vv

vv0

vv

tt

∆v = v-v0

∆t = t-t0 = t

tt


18


xx

xx0

xx

tt

∆x = x-x0

∆t = t-t0 = t

t =0t =0

xx = x0+ <vv >t

Substitute:

<v v > = (vv0 +vv)/2

xx = xx0+ [(vv0 +vv)/2]t

x x = x0+ ½(vv0t +vv t)

x x = x0+ ½(vv0t )+½(vv0+aat)(t)

x x = x0+vv0t +½aat2


19

Kinematic Equations of Motion For Constant Acceleration

vv = vv0+aa t » t = (vv - vv0)/aa

x x = x0+vv0t +½a a t2

Substitute for t

x x = x0+vv0 (vv - vv0)/aa +½aa[(vv - vv0)/aa]2

Simplify

(vv2 - vv02) = 2aa(x x -- x0)


20

Kinematic Equations of Motion For Constant Acceleration

vv = vv0+aa t

x x = x0+vv0t +½a a t2

(vv2 - vv02) = 2aa(x x -- x0)


21

Falling Objects:Uniform (constant)Acceleration

Acceleration due to gravity: g = 9.80m/s2 = 32.0ft/s2

• Galileo studied falling objects

– Postulated: All objects fall with same constant acceleration in absence of air or other resistance

– Predicted: Free falling object travels a distance, y, proportional to time squared (t2)

y ~ t2

– Derived: Free fall equation

y y = y0+vv0t +½g g t2


22

Falling Objects:Uniform (constant)Acceleration

No air resistance– In a vacuum– Feathers and boulders

fall with same acceleration ~ g

– x ~ t2

= t == t = ∆ ∆yy


23

Graphical Analysis of Linear Motion

Linear Equation: (y = mx + b)

• m is slope

• b is y intercept

Eq of Motion for velocity

• x = x0 + vt

Slope: v = ∆x/∆t

Y intercept: x0

xx

tt

∆x = x-x0

∆t = tx0


24

Graphical Analysis of Linear Motion

Eq of Motion for position

• v = v0 + at

Slope: a = ∆v/∆t

y intercept: v0tt

∆v = v - v0

∆t = tvv00

vv

September 15, 2009Physics I Lesson 2 Dr J. Tison 1 Kinematics: One Dimensional Motion.

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Transcript of September 15, 2009Physics I Lesson 2 Dr J. Tison 1 Kinematics: One Dimensional Motion.