General Physics IGeneral Physics I

Lecture 1: Motion in 1DLecture 1: Motion in 1D

Prof. WAN, Xin （万歆）

xinwan@zju.edu.cnhttp://zimp.zju.edu.cn/~xinwan/

Motion in One DimensionMotion in One Dimension

DefinitionsDefinitions

● Displacement

● Average velocity

● Average speed

x f xi

if xxΔx

x

if

ifx tt

xx=

Δt

Δxv

Δ x

timetotal

distancetotal=

A Round Trip to ShanghaiA Round Trip to Shanghai

• Suppose the two train stations (Hangzhou & Shanghai) are 150 km apart. It takes 1 hour and 15 minutes to go to Shanghai, but 45 minutes to come back.

– Average velocity?

– Average speed?

• Velocity and speed NOT interchangeable in physics!

A Figure is Worth 1,000 WordsA Figure is Worth 1,000 Words

t

x

Hangz

hou

Shangh

ai

Jiaxin

g

1 hour 2 hours

More Like RealityMore Like Reality

t

x

Hangz

hou

Shangh

ai

Jiaxin

g

1 hour 2 hours

Instantaneous VelocityInstantaneous Velocity

t

x

Hangz

hou

Shangh

ai

Jiaxin

g

1 hour 2 hours

Δ t

Δ x

vx≡limΔ t→0

Δ xΔ t

A

B

Tangent Line; DerivativeTangent Line; Derivative

1)( nnAtdt

tdxA

B dt

dx=

Δt

Δxv

Δtx

0lim

derivative of x with respect to t

nAttx )(Example:

Average AccelerationAverage Acceleration

t

x

t

vx

Δvx

Δ t

ax≡Δvx

Δ t=v xf−vxi

t f−ti

vxf

vxi

(Instantaneous) Acceleration(Instantaneous) Acceleration

dt

dv=

Δt

Δva xx

tx

0lim

2

2

dt

xd

dt

dx

dt

d

dt

dv=a x

x

Example 2.4Example 2.4

The velocity of a particle moving along the x axis varies in time according to the expression

Where t is in seconds. (a) Determine the acceleration at t = 2.0 s.

m/s,)540( 2tvx

tdt

dv=

Δt

Δva xx

tx 10lim

0

Example 2.4Example 2.4

If you are not familiar with calculus, don’t worry. Just take the limit yourself. Take t = 2 s.

22 )2540 tttt

2510 tttvx

2)(540)( ttttvx

2

0m/s 2010lim

t

Δt

Δva x

tx

)540()( 2ttvx

The unit is important!

Acceleration can be negative!

Example 2.4Example 2.4

(b) Find the average acceleration in the time interval t = 0 to t = 2.0 s.

if

xixfxx tt

vv=

Δt

Δva

Recall

We need velocities at t = 0 and 2.0 s.

2m/s 10xa

UnitsUnits

• Basic quantities– Length (L)– Mass (M)– Time (T)

• The basic SI system of units– meter (m)– kilogram (kg)– second (s)

• Prefixes: km, mm, m, nm, g, …

We say that the dimension of distance is length.

L/T][ v

Dimension of velocity:

If you convert everything to the basic SI units of meter, kilogram, and second, you can omit them during the calculation, but put back the correct unit at the end.

Position-time graph

Velocity-time graph

Acceleration-time graph

?

IntegrationIntegration

f

in

t

t

xn

nxnt

if dttvtvxxx )(lim0

Displacement = area under the vx(t) curve

Kinematic EquationsKinematic Equations

dtadvdt

dv=a xx

xx

t

t

xxix

i

dttav=tv ')'()(

1)(, Ctvvtt ixxii

1')'()( Cdtta=tvt

t

xx

i

Kinematic EquationsKinematic Equations

dtvdxdt

dx=v xx

'

)"("')(')'()(t

t

x

t

t

ixii

t

t

xi

iii

tadtdtttvxdttvx=tx

2)(, Ctxxtt iii

2')'()( Cdttv=txt

t

x

i

Example: Constant AccelerationExample: Constant Acceleration

Constant AccelerationConstant Acceleration

xv

2xfxi

x

vv=v

xiv

tavvttt xxixffi ,,0Set

t

xfv

0

Constant AccelerationConstant Acceleration

xv tvvxx xfxiif 2

1

xiv

tavvttt xxixffi ,,0Set

t

xfv

0

2

2

1tatv xxi

2

tax

2

tax

x

xixf

a

vv

2

22

Dimension AnalysisDimension Analysis

tvvxx xfxiif 2

1

2

2

1tatv xxi

x

xixf

a

vv

2

22

LTT

L][ vt

LTT

L][ 2

22 at

LL/T

(L/T)2

22

a

v

Leaning Tower of PisaLeaning Tower of Pisa

If Galileo did perform the legendary experiment, why did he pick the Leaning Tower of Pisa?

What can we do today?

In ActionIn Action

Read DataRead Data

t x y

0 0 0

1 2.5 0

2 5 -0.7

3 7.8 -1.6

4 10.2 -3

5 12.8 -4.5

6 15.2 -6.4

7 17.7 -9

Data Analysis IData Analysis I

t

x

Data Analysis IIData Analysis II

t

y

Data Analysis IIIData Analysis III

t2

y

Significant FiguresSignificant Figures

• Physical quantities measured are known only to within the limits of the experimental uncertainty. – Quality of the apparatus

– Skill of the experimenter

– Number of measurements performed

• Significant figures– (5.5 ±0.1) cm (6.4 ± 0.1) cm = (35 ± 1) cm2

• Zero may be significant– 1.5 kg ≠ 1.50 kg ≠ 1.500 kg

QuestionQuestion

tvvxx xfxiif 2

1change? equations

following thedo how,)(Suppose cttax

2

2

1tatv xxi

x

xixf

a

vv

2

22

Results for = 0