Physics 100: Lecture 1
Agenda for Today




Advice
Scope of this course
Measurement and Units
Fundamental units
Systems of units
Converting between systems of units
Dimensional Analysis
1-D Kinematics (review)
Average & instantaneous velocity and acceleration
Motion with constant acceleration
Physics 111: Lecture 1, Pg 1
Course Info & Advice

See info on the World Wide Web (heavily used in Physics 101)
Go to http://www.physics.uiuc.edu and follow “courses” link
to the Physics 101 homepage

Course has several components:
Lecture: (me talking, demos and Active learning)
Discussion sections (group problem solving)
Homework sets, Web based
Labs: (group exploration of physical phenomena)
If you miss a lab or discussion you should always try to make
it up as soon as possible in another section!!

The first few weeks of the course should be review, hence the
pace is fast. It is important for you to keep up!
Physics 111: Lecture 1, Pg 2
Lecture Organization

Three main components:
Lecturer discusses class material
» Follows lecture notes very closely
Lecturer does as many demos as possible
» If you see it, you gotta believe it!
» Look for the symbol
Students work in groups on conceptual
“Active Learning” problems
» Usually three per lecture
Physics 111: Lecture 1, Pg 3
Scope of Physics 101

Classical Mechanics:
Mechanics: How and why things work
Classical:
» Not too fast
(v << c)
» Not too small (d >> atom)

Most everyday situations can be described in these terms.
Path of baseball
Orbit of planets
etc...
Physics 111: Lecture 1, Pg 4
Units


How we measure things!
All things in classical mechanics can be expressed in terms
of the fundamental units:
Length
Mass
Time

L
M
T
For example:
Speed has units of L / T (i.e. miles per hour).
Force has units of ML / T2 etc... (as you will learn).
Physics 111: Lecture 1, Pg 5
Length:
Distance
Radius of visible universe
To Andromeda Galaxy
To nearest star
Earth to Sun
Radius of Earth
Sears Tower
Football field
Tall person
Thickness of paper
Wavelength of blue light
Diameter of hydrogen atom
Diameter of proton
Length (m)
1 x 1026
2 x 1022
4 x 1016
1.5 x 1011
6.4 x 106
4.5 x 102
1.0 x 102
2 x 100
1 x 10-4
4 x 10-7
1 x 10-10
1 x 10-15
Physics 111: Lecture 1, Pg 6
Time:
Interval
Age of universe
Age of Grand Canyon
32 years
One year
One hour
Light travel from Earth to Moon
One cycle of guitar A string
One cycle of FM radio wave
Lifetime of neutral pi meson
Lifetime of top quark
Time (s)
5 x 1017
3 x 1014
1 x 109
3.2 x 107
3.6 x 103
1.3 x 100
2 x 10-3
6 x 10-8
1 x 10-16
4 x 10-25
Physics 111: Lecture 1, Pg 7
Mass:
Object
Milky Way Galaxy
Sun
Earth
Boeing 747
Car
Student
Dust particle
Top quark
Proton
Electron
Neutrino
Mass (kg)
4 x 1041
2 x 1030
6 x 1024
4 x 105
1 x 103
7 x 101
1 x 10-9
3 x 10-25
2 x 10-27
9 x 10-31
1 x 10-38
Physics 111: Lecture 1, Pg 8
Units...

SI (Système International) Units:
mks: L = meters (m), M = kilograms (kg), T = seconds (s)
cgs: L = centimeters (cm), M = grams (gm), T = seconds (s)

British Units:
Inches, feet, miles, pounds, slugs...

We will use mostly SI units, but you may run across some
problems using British units. You should know how to convert
back & forth.
Physics 111: Lecture 1, Pg 9
Converting between different systems of units

Useful Conversion factors:
1 inch
= 2.54 cm
1 m
= 3.28 ft
1 mile
= 5280 ft
1 mile
= 1.61 km

Example: convert miles per hour to meters per second:
1
mi
mi
ft
1 m
1 hr
m
1
 5280


 0.447
hr
hr
mi 3.28 ft 3600 s
s
Physics 111: Lecture 1, Pg 10
Dimensional Analysis

This is a very important tool to check your work
It’s also very easy!

Example:
Doing a problem you get the answer distance
d = vt 2 (velocity x time2)
Units on left side = L
Units on right side = L / T x T2 = L x T

Left units and right units don’t match, so answer must be
wrong!!
Physics 111: Lecture 1, Pg 11
Lecture 1, Act 1
Dimensional Analysis

(a)
The period P of a swinging pendulum depends only on
the length of the pendulum d and the acceleration of
gravity g.
Which of the following formulas for P could be
correct ?
P = 2 (dg)2
(b)
P  2
d
g
(c)
d
P  2
g
Given: d has units of length (L) and g has units of (L / T 2).
Physics 111: Lecture 1, Pg 12
Lecture 1, Act 1
Solution


Realize that the left hand side P has units of time (T )
Try the first equation
2
4
L
L


(a)  L 
  4 T
 T2 
T
(a)
P  2 dg 
2
(b)
Not Right !!
P  2
d
g
(c)
d
P  2
g
Physics 111: Lecture 1, Pg 13
Lecture 1, Act 1
Solution

Try the second equation
(b)
(a)
L
 T2  T
L
T2
P  2 dg 
2
Not Right !!
(b)
P  2
d
g
(c)
d
P  2
g
Physics 111: Lecture 1, Pg 14
Lecture 1, Act 1
Solution

Try the third equation
(c)
(a)
L
 T2 T
L
T2
P  2 dg 
2
(b)
This has the correct units!!
This must be the answer!!
P  2
d
g
(c)
d
P  2
g
Physics 111: Lecture 1, Pg 15
Motion in 1 dimension

In 1-D, we usually write position as x(t1 ).

Since it’s in 1-D, all we need to indicate direction is + or .

Displacement in a time t = t2 - t1 is
x = x(t2) - x(t1) = x2 - x1
x
x
some particle’s trajectory
in 1-D
x
2
x
1
t1
t
t2
t
Physics 111: Lecture 1, Pg 16
1-D kinematics


Velocity v is the “rate of change of position”
Average velocity vav in the time t = t2 - t1 is:
v av 
x( t 2 )  x( t1 ) x

t 2  t1
t
x
x
trajectory
x
2
Vav = slope of line connecting x1 and x2.
x
1
t1
t2
t
t
Physics 111: Lecture 1, Pg 17
1-D kinematics...


Consider limit t1 t2
Instantaneous velocity v is defined as:
v( t ) 
x
x
dx( t )
dt
so v(t2) = slope of line tangent to path at t2.
x
2
x
1
t1
t2
t
t
Physics 111: Lecture 1, Pg 18
1-D kinematics...


Acceleration a is the “rate of change of velocity”
Average acceleration aav in the time t = t2 - t1 is:
aav 

v ( t 2 )  v ( t1 ) v

t 2  t1
t
And instantaneous acceleration a is defined as:
dv ( t ) d 2 x( t )
a( t ) 

dt
dt 2
using v ( t ) 
dx( t )
dt
Physics 111: Lecture 1, Pg 19
Recap

If the position x is known as a function of time, then we can
find both velocity v and acceleration a as a function of time!
x
x  x( t )
dx
dt
dv
d 2x
a 

dt
dt 2
v 
v
a
t
t
t
Physics 111: Lecture 1, Pg 20
More 1-D kinematics


We saw that v = dx / dt
In “calculus” language we would write dx = v dt, which we
can integrate to obtain:
t2
x (t 2 )  x (t1 )   v (t )dt
t1

Graphically, this is adding up lots of small rectangles:
v(t)
+ +...+
= displacement
t
Physics 111: Lecture 1, Pg 21
1-D Motion with constant acceleration

High-school calculus:  t n dt 

Also recall that a 
dv
dt
1
t n 1  const
n 1

Since a is constant, we can integrate this using the above
rule to find:
v   a dt  a  dt  at  v 0

Similarly, since v 
dx
we can integrate again to get:
dt
1
x   v dt   ( at  v 0 )dt  at 2  v 0 t  x0
2
Physics 111: Lecture 1, Pg 22
Recap

So for constant acceleration we find:
Plane
w/ lights
x
1
x  x0  v 0 t  at 2
2
v  v 0  at
a  const
v
a
t
t
t
Physics 111: Lecture 1, Pg 23
Lecture 1, Act 2
Motion in One Dimension

When throwing a ball straight up, which of the following is
true about its velocity v and its acceleration a at the
highest point in its path?
(a) Both v = 0 and a = 0.
(b) v  0, but a = 0.
y
(c) v = 0, but a  0.
Physics 111: Lecture 1, Pg 24
Lecture 1, Act 2
Solution

Going up the ball has positive velocity, while coming down
it has negative velocity. At the top the velocity is
momentarily zero.
x

Since the velocity is
continually changing there must
v
be some acceleration.
In fact the acceleration is caused
by gravity (g = 9.81 m/s2).
(more on gravity in a few lectures) a

The answer is (c) v = 0, but a  0.
t
t
t
Physics 111: Lecture 1, Pg 25
Useful Formula
v  v 0  at

Solving for t:
t
v  v0
a
1
x  x0  v 0 t  at 2
2

Plugging in for t:
 v  v0  1  v  v0 
x  x0  v 0 
  a

 a  2  a 
2
v 2  v 0  2a( x  x0 )
2
Physics 111: Lecture 1, Pg 26
Alternate (Calculus-based) Derivation
dv dv dx
(chain rule)
a  
dt dx dt
dv
a  v
dx
x

x
a  dx  v  dv
v
 a dx  a  dx   v  dv
x0
x0
(a = constant)
v0
1 2 2
 a ( x - x0 )  ( v  v 0 )
2
v 2  v 0  2a( x  x0 )
2
Physics 111: Lecture 1, Pg 27
Recap:

For constant acceleration:
Washers
1
x  x0  v 0 t  at 2
2
v  v 0  at
a  const

From which we know:
v 2  v 02  2a(x  x0 )
v av
1
 (v 0  v)
2
Physics 111: Lecture 1, Pg 28
Problem 1

A car is traveling with an initial velocity v0. At t = 0, the
driver puts on the brakes, which slows the car at a rate of
ab
vo
ab
x = 0, t = 0
Physics 111: Lecture 1, Pg 29
Problem 1...

A car is traveling with an initial velocity v0. At t = 0, the
driver puts on the brakes, which slows the car at a rate of
ab. At what time tf does the car stop, and how much farther
xf does it travel?
v0
ab
x = 0, t = 0
v=0
x = x f , t = tf
Physics 111: Lecture 1, Pg 30
Problem 1...

Above, we derived: v = v0 + at

Realize that a = -ab

Also realizing that v = 0 at t = tf :
find 0 = v0 - ab tf or
tf = v0 /ab
Physics 111: Lecture 1, Pg 31
Problem 1...

To find stopping distance we use:
v 2  v 02  2a(x  x0 )

In this case v = vf = 0, x0 = 0 and x = xf
 v 0  2( ab )xf
2
2
v
xf  0
2 ab
Physics 111: Lecture 1, Pg 32
Problem 1...



So we found that
2
1 v0
tf 
, xf 
ab
2 ab
v0
Suppose that vo = 65 mi/hr = 29 m/s
Suppose also that ab = g = 9.81 m/s2
 Find that tf = 3 s and xf = 43 m
Physics 111: Lecture 1, Pg 33
Tips:

Read !
Before you start work on a problem, read the problem
statement thoroughly. Make sure you understand what
information is given, what is asked for, and the meaning
of all the terms used in stating the problem.

Watch your units !
Always check the units of your answer, and carry the
units along with your numbers during the calculation.

Understand the limits !
Many equations we use are special cases of more
general laws. Understanding how they are derived will
help you recognize their limitations (for example,
constant acceleration).
Physics 111: Lecture 1, Pg 34
Recap of today’s lecture

Scope of this course
Measurement and Units (Chapter 1)
Systems of units
(Text: 1-1)
Converting between systems of units
(Text: 1-2)
Dimensional Analysis
(Text: 1-3)
1-D Kinematics
(Chapter 2)
Average & instantaneous velocity
and acceleration
(Text: 2-1, 2-2)
Motion with constant acceleration
(Text: 2-3)
Example car problem
(Ex. 2-7)

Look at Text problems Chapter 2: # 49, 54, 71, 122



Physics 111: Lecture 1, Pg 35