1
Example:
EVERYTHING MOVES!!!!!
THERE IS NO APSOLUTE REST!!!!
 A frame of reference – is a perspective from which a
system is observed together with a coordinate system used to
describe motion of that system.
Classical Mechanics: deals with the physical laws describing the
motion of bodies under the action of a system of forces.
It successfully describes the motion for object that are
1. large compared with the dimensions of atoms (10-10 m)
2. moving at speeds that are small compared to the speed
of light (3x108 m/s)
Kinematics – is the branch of classical mechanics that describes
the motion of objects and systems (groups of objects) without
consideration of the forces acting on them. Motion is described
in terms of distance/displacement, speed/velocity, and
acceleration.
Dynamics – is the study of forces explaining why objects change
the velocity.
The movement of an object through space can be quite
complex.
There can be internal motions, rotations, vibrations, etc…
This is the combination of rotation
(around its center of mass) and the
motion along a line - parabola.
f we treat the hammer as a particle
the only motion is translational
motion (along a line) through space.
A racing car travels round a circular track of radius 100 m.
The car starts at O. When it has travelled to P its displacement as
measured from O is
A
B
C
D
100 m due East
100 m due West
100 √2 m South East 
100 √2 m South West
 Vectors and Scalars
Each physical quantity will be either a scalar or a vector.
 Scalar is a quantity which is fully described by a magnitude
(or numerical value with appropriate unit) alone.
Temperature, length, mass, time, speed, …
 Vector is a quantity which is fully described by both
magnitude (or numerical value and unit) and direction.
Displacement, velocity, force,…
Scalar
distance - 50 km
speed - 70 km h-1
Vector
displacement: 50 km, E
velocity:
70 km h-1, S-W
scalars obey the rules of ordinary algebra:
2 kg of potato + 2 kg of potato = 4 kg of potato
Vectors do not obey the rules of ordinary algebra:
2 Newtons + 4 Newtons is not necessarily 6 Newtons
Kinematics in One Dimension
Our objects will be represented as point objects (particles) so
they move through space without rotation.
Simplest motion: motion of a particle along a line – called:
translational motion or one-dimensional (1-D) motion.
The sum of two (or more) vectors
depends on their directions!!!!
 Displacement is the shortest distance in a given direction.
It tells us not only the
distance of an object from
initial position but also the
direction from that point.
(it is the change in position)
Example: an ant wanders from P to Q position
distance traveled = 1 m
shortest distance from P to Q = 0.4 m
displacement of the ant = 0.4 m, SE
Example:
1) x1 = 7 m, x2 = 16 m
2 N,West + 2N,East = 0
 Average and Instantaneous Velocity
in some time
interval
when we say velocity we
mean instantaneous velocity
 DEF:
Average velocity is the displacement divided by elapsed time.
v avg =
x3 = 12 m ∆x = 5 m
“+” direction & distance
at one
instant
displacement
time
v avg =
x 2 -x1
Δx
=
t 2 -t1
Δt
SI unit : m/s
(it obviously has direction, the same as displacement)
 Instantaneous velocity
2) x1 = 7 m, x2 = 2 m
∆x = –5m
“–” direction & distance
When one is looking for instantaneous velocity all one is looking
for is the velocity at a particular moment in time. It is like asking,
“When time equals 5s how fast is the object going?”
2
The instantaneous velocity is velocity at a given moment. It tells
us how fast (speedometer) and in what direction is the object
moving right now. So we simply take the equation above and
assume that Δt is very, very small .
In math we say time interval Δt approaches zero, and we write:
Example:
A racing car travels round a circular track of
radius 100 m. The car starts at O.
It travels from O to P in 20 s.
Its velocity was 10 m/s.
Its speed was πr/t = 16 m/s.
lim Δx
dx
v=
=
Δt  0 Δt
dt
The car starts at O. It travels from O back to O in 40 s.
If an object is traveling at a constant velocity than its
instantaneous velocity is the same at all times. But, if the object
is accelerating (speeding up or slowing down) its velocity at any
particular second is different than its velocity at any other
second. So, we will need an equation that will tell us how fast an
object is going at a particular time while it is accelerating.
 Geometrical Interpretation of Average
and Instantaneous Velocity
Let’s assume that we know position at any time → graph.
slope =
vavg
Δx
= v avg
Δt
Average velocity of a particle
during the time interval Δt is
equal to the slope of the
straight line joining the initial
(P) and final (T) position on the
position-time graph.
gives us no details of the motion between
initial and final points.
Instantaneous velocity of a particle at some position P at
time t is equal to the slope of the tangent line at P on the
position-time graph.
lim Δx dx

=v
Δt  0 Δt dt
 Average and Instantaneous Speed
How fast do your eyelids move when you blink? Displacement is
zero, so vavg = 0. How fast do you drive in one hour if you drive
Its velocity was 0 m/s. Its speed was 2πr/t = 16 m/s.
 Let’s look at the motion with constant velocity
so called uniform motion
in that case, velocity is the same at all times so v = vavg
at all times, therefore:
x or x = vt
t
Object moving at constant velocity covers the same distance in
the same interval of time.
v=
 Average and Instantaneous Acceleration
 DEF: Acceleration is the change in velocity per unit time
a=
Δv
Δt
(Change in velocity ÷ time taken)
it has direction, the same as the change of velocity
m
SI unit:  a  = s = m/s2
s
a = 3 m/s2 means that velocity changes 3 m/s every second!!!
If an object’s initial velocity is 4 m/s then after one second it will
be 7 m/s, after two seconds 10 m/s,…
 Instantaneous acceleration is the change in velocity over an
infinitesimal time interval.
a=
lim Δv
dv
=
Δt  0 Δt
dt
zigzag? Displacement is shorter than distance traveled. To get
the answers to these questions we introduce speed:
 Geometrical Interpretation of Average
and Instantaneous Acceleration
Average speed - during some time interval
If we know object’s velocity as a function of time, we can find
both average and instantaneous acceleration.
 Speed is the distance object covers per unit time.
v avg =
distance travelled
Δt
it tells us how fast the object is moving
on the other hand velocity tells us how fast and in what
direction object would be moving if it covered the shortest
distance from beginning to the end point in the same amount of
time.
if motion is 1-D without changing direction:
average speed = magnitude of average velocity because
distance traveled = displacement
• instantaneous speed = magnitude of instantaneous velocity
In velocity – time graph
average acceleration during a
time interval Δt is the slope of a
straight line joining the initial
and final velocity on the
velocity- time graph,
and
instantaneous acceleration at
some time is the slope of the line tangent to the v vs. t curve at
that time.
3
 Geometrical Interpretation of the Area under Graph
Acceleration can cause: 1. speeding up 2. slowing down
3. and/or changing direction
the area under a velocity –
time graph gives the
displacement covered in time
(t2 – t1)
the area under acceleration
– time graph is the change in
velocity in time interval
(t2 – t1) .
 Uniformly Accelerated Motion
motion with constant acceleration
v = u + at
example:
u = 2 m/s
a = 3 m/s
→
velocity v at any time t = initial velocity u
increased by a, every second
t (s)
0
1
2
3
𝑣𝑎𝑣𝑔 =
𝑣−𝑢
𝑡
v (m/s)
2
5
8
11
speed increases 3 m/s
EVERY second.
← arithmetic sequence, so:
(2 + 5 + 8 + 11)𝑚/𝑠
(2 + 11)𝑚/𝑠
=
= 6.5 𝑚/𝑠
4
2
 In general: for the motion with constant acceleration:
v avg =
u+ v
2
𝒗𝒂𝒗𝒈 =
x = vavg t =
𝒖+𝒗
𝟐
t
𝒙 = 𝒖𝒕 +
• 2. if velocity and acceleration (change in velocity) are in the
opposite directions, speed of the body is decreasing.
• 3. If an object changes direction even at constant speed it
is accelerating. Why? Because the direction of the car is
changing and therefore its velocity is changing. If its velocity
is changing then it must have acceleration.
A stone is rotating around the center of a circle.
The speed is constant, but velocity is not –
direction is changing as the stone travels around,
therefore it must have acceleration.
Velocity is tangential to the circular path at any time.
ACCELERATION IS ASSOCIATED WITH A FORCE!!!
The force (provided by the string) is forcing the stone to move in
a circle giving it acceleration perpendicular to the motion –
toward the CENTER OF THE CIRCLE - along the force. This is the
acceleration that changes velocity by changing it direction only.
When the rope breaks, the stone goes off in the tangential
straight-line path because no force acts on it.
In the case of moon acceleration is caused by gravitational force
between the earth and the moon. So,
acceleration is always toward the earth.
That acceleration is changing velocity
(direction only).
1. weakening gravitational force would
result in the moon getting further and
further away still circling around earth.
2. no gravitational force all of a sudden: there wouldn’t be
acceleration – therefore no changing the velocity (direction) of
the moon, so moon flies away in the direction of the velocity at
that position ( tangentially to the circle).
 Uniform Accelerated Motion equations
v = u + at
• 1. if velocity and acceleration (change in velocity) are in the
same direction, speed of the body is increasing.
Examples of changing direction only:
let:
t = the time interval for which the body accelerates
a = constant acceleration
u = the velocity at time t = 0, the initial velocity
v = the velocity after time t, the final velocity
x = the displacement covered in time interval t
1. from the definition of a: 𝑎 =
So beware: both velocity and acceleration are vectors. Therefore
𝒖+𝒗
𝟐
𝒂
𝟐
𝒕𝟐
𝒗𝟐 = 𝒖𝟐 + 𝟐𝒂𝒙
In addition to these equations to solve a problem with constant
acceleration you’ll need to introduce your own coordinate
system, because displacement, velocity and acceleration are
vectors (they have directions).
3. The moon has no speed – it moves toward the earth –
accelerated motion in the straight line – crash
4. High speed – result the same as in the case of weakening
gravitational force
Only the right speed and acceleration (gravitational force) would
result in circular motion!!!!!!!
4
 Free Fall
Free fall is vertical (up and/or down) motion of a body
where gravitational force is the only or dominant force
acting upon it
Gravitational force gives all bodies regardless of mass or shape,
when air resistance can be ignored, the same acceleration.
This acceleration is called free fall or gravitational acceleration
(symbol g – due to gravity).
Free fall acceleration at Earth’s surface is about 9.8 m/s2 toward
the center of the Earth.
Equations, substitution, solution:
1. at the top:
v = u + gt = 0
20 – 10t = 0
t=2s
free fall is symmetrical up and down
t = 4 s for the whole motion
2. max height 𝑦𝑚𝑎𝑥
𝑦𝑚𝑎𝑥 =
𝑢+𝑣
𝑡
2
𝑦𝑚𝑎𝑥 = 20 m
Let’s throw an apple equipped with a speedometer
upward with some initial speed u.
That means that apple has velocity u as it leaves our hand.
The speed would decrease by 9.8 m/s every second on the way
up, at the top it would reach zero, and increase by 9.8 m/s for
each successive second on the way down.
You would get the same result if you chose y axis to be positive
downwards. In that case
g = 10 m/s/s, u = – 20 m/s.
t = 2 s to the top (–20 + 10t = 0)
g depends on how far an object is from the center of the Earth.
The farther the object is, the weaker the attractive gravitational
force is, and therefore the gravitational acceleration is smaller.
At the bottom of the valley you accelerate faster (slightly)
then on the top of the Himalayas.
Gravitational acceleration at the distance 330 km from the
surface of the Earth (where the space station is) is ̴7.8 m/s2.
In reality – good vacuum (a container with the air pumped out)
can mimic ideal free fall.
August 2, 1971 experiment was conducted on the Moon – David
Scott simultaneously released geologist’s hammer and falcon’s
feather. Falcon’s feather dropped like the hammer. They
touched the surface at the same time.
 Equations are the ones for uniform accelerated
motion with a = g
𝑣𝑎𝑣𝑔 =
v = u + gt
𝑦 = 𝑣𝑎𝑣𝑔 𝑡 =
𝑢+𝑣
𝑡
2
2
𝑡=
−20+0
2
𝑡2
𝑣 2 = 𝑢2 + 2𝑔𝑦
2. Mr. Rutzen, hovering in a helicopter 200 m above our school
suddenly drops his pen.
How much time will the students have to save themselves? What
is the velocity/speed of the pen when it reaches the ground?
Givens:
u = 0 m/s (dropped)
g = 10 m/s2
𝑎
2
𝑡2
200 = 0 + 5 t2
t = 6.3 s
at the top: v = 0 (object is changing direction)
v = u + gt
v = 0 + 10 t
Very often we will use: g = 10 m/s2 (simplicity)
v = 63 m/s (speed)
velocity is 63 m/s downward
To describe the motion we need coordinate system.
EXAMPLES:
1. Dr. Huff, a very strong lady, throws a ball upward with initial
speed of 20 m/s.
How high will it go? How long will it take for the ball to come
back?
Givens:
Unknowns:
u = 20 m/s
t=?
g = - 10 m/s2
y=?
at the top v = 0
2 = −20 𝑚
negative sign in the coordinate system in which downward is
positive means that 𝑦𝑚𝑎𝑥 is upward.
y = 𝑢𝑡 +
𝑎
2
𝑢+𝑣
Unknowns:
t=?
v=?
𝑢+𝑣
2
𝑦 = 𝑢𝑡 +
𝑦𝑚𝑎𝑥 =
20 – 10t = 0
3. Mrs. Radja descending in a balloon at the speed of 5 m/s
above our school drops her car keys from a height of 100 m.
How much time will the students have to save themselves?
What is the velocity of the keys when they reach the ground?
t=?
v=?
y = 𝑢𝑡 +
𝑎
5
𝑡2
2
𝑎
as you can see equation 𝑦 = 𝑢𝑡 + 𝑡
2
takes care of direction of initial velocity
100 = 5𝑡 + 5 𝑡 2
5 𝑡 2 + 5𝑡 − 100 = 0
𝑡 2 + 𝑡 − 20 = 0
𝑡=
− 1 ±9
2
t=4s
2
d.
time can be only positive
v = u + gt
v = 20 – 10 x 6 = – 40 m/s
(direction : velocity)
speed at the bottom is 40 m/s
still enough time to save yourself – start running
v = u + gt
v = 50 m/s, downwards
it is dangerous speed; run away as fast as you can
4. Dr. Huff, our very strong lady, goes to the roof and throws a
ball upward. The ball leaves her hand with speed 20 m/s.
Ignoring air resistance calculate
a. the time taken by the stone to reach its maximum height
b. the maximum height reached by the ball.
c. the height of the building is 60 m. How long does it take for
the ball to reach the ground?
d. what is the speed of the ball as it reaches the ground?
Graphs of free fall motion
v =gt
v = 10 t
Time
(s)
0
1
2
3
4
a. at the top: v = u + gt = 0
20 – 10t = 0
we ignore the height of any human being
b. maximum height:
𝑦𝑚𝑎𝑥 = 𝑢𝑡 +
or
𝑦𝑚𝑎𝑥 =
𝑎
2
𝑡 2 = 20 𝑚
Velocity
(m/s)
0
10
20
30
40
y=
g 2
t
2
y = 5 t2
Distance
(m)
0
5
20
45
60
t=2s
constant slope –
constant acceleration
𝑢+𝑣
𝑡 = 20 𝑚
2
c. y = - 60 m
𝑎
𝑦 = 𝑢𝑡 +
2
𝑡2
– 60 = 20 t – 5 t2
t2 – 4 t – 12 = 0
t=2±4
as the can not e negative t = 6 s
or
from the top
𝑦 = 𝑢𝑡 +
u=0
𝑎
2
𝑡2
– 80 = – 5 t2
t=4s
t=4+2=6s
y = - 80 m
changing slope –
changing speed → acceleration
6
If air resistance can not be neglected, there is additional force
(drag force) acting on the body in the direction opposite to
velocity.
terminal velocity is maximum velocity an object can reach in
air/any fluid.
Acceleration is getting smaller due to air resistance and
eventually becomes zero.
When the force of the air resistance equals gravity, the object
will stop accelerating and maintain the same speed.
It is different for different bodies.
Air Drag and Terminal Velocity
If a raindrops start in a cloud at a height h = 1200m above the
surface of the earth they hit us at 340mi/h; serious damage
would result if they did. Luckily: there is an air resistance
preventing the raindrops from accelerating beyond certain
speed called terminal speed….
How fast is a raindrop traveling when it hits the ground?
It travels at 7m/s (17 mi/h) after falling approximately only 6 m.
This is a much “kinder and gentler” speed and is far less
damaging than the 340mi/h calculated without drag.
The terminal speed for a skydiver is about 60 m/s
(pretty terminal if you hit the deck)