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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 (3x10 8 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. Explains motion and causes of changes using
concepts of force and energy.
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.
If we treat the hammer as a particle the only motion is translational motion (along a line) through
space.
Kinematics in One Dimension (along a line)
Our objects are represented as point objects (particles) so they move through space without rotation around its center of
mass.
Simplest motion: motion of a particle along a line – called:
translational motion or one-dimensional (1-D) motion.
 Displacement of an object is the shortest distance from its
initial to the final position. It is a vector !!!!!!!!!!!
.
The displacement tells us how far an object is from
its starting position and in what direction.
Example:
1) x1 = 7 m, x2 = 16 m
2) x1 = 7 m, x2 = 2 m
x3 = 12 m
∆x = 5 m (“+”= direction; distance = 5m)
∆x = –5m
(“–” = direction; distance = 5m)
Example:
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
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 Average and Instantaneous Velocity
in some time
interval
at one
instant
when we say velocity we mean
instantaneous velocity
 Average velocity is the displacement divided by elapsed 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
Instantaneous velocity is the velocity at one instant. THERE IS NO GENERAL FORMULA EXCEPT THROUGH DERIVATIVES.
The speedometer of a car reveals information about the instantaneous speed of your car. It shows your speed at a particular
instant in time. If direction is included you have instantaneous velocity.
 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 zigzag and the magnitude of the displacement is different from distance?
To get the answers to these questions we introduce speed:
 Average velocity is the displacement divided by elapsed time.
v avg =
distance travelled
Δt
it tells us how fast the object is moving
KCR train has travelled a distance of about 6.7 km from University Station
to Tai Po Market Station. But if we measure their separation by drawing
a straight line on the map, we will find that Tai Po Market Station is
only 5.4 km from University Station, roughly in the North-West direction.
This is the displacement of the train.
We take the KCR trip from University Station to Tai Po Market Station.
It takes about 6 minutes to travel a distance of 6.7 km. Thus,
6.7
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑎𝑖𝑛 =
= 1.12 𝑘𝑚/𝑚𝑖𝑛
6
The displacement of Tai Po Market Station from University Station is 5.4 km, so
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑎𝑖𝑛 =
5.4
= 0.9 𝑘𝑚/ min 𝑖𝑛 𝑡ℎ𝑒 𝑛𝑜𝑟𝑡ℎ − 𝑤𝑒𝑠𝑡 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛.
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This is smaller than the average speed of the train.
So why do we care of velocity at all? OK, it gives us direction what is very important (just imagine airplane controller with
information only on speed of airplanes not on directions). But we saw that average speed is greater in general than
magnitude of average velocity. So why is concept of velocity so important?
Because acceleration is a vector, and all equations are actually vector equations.
Acceleration can be in any direction to the velocity and the motion will depend on that.
ONLY: if motion is 1-D without changing direction:
• average speed = magnitude of average velocity because distance traveled = magnitude of displacement
• instantaneous speed = magnitude of instantaneous velocity
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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, S. Its speed was πr/t = 16 m/s.
The car starts at O. It travels from O back to O in 40 s.
Its velocity was 0 m/s. Its speed was 2πr/t = 16 m/s.
 Motion with constant velocity – uniform motion
in that case, velocity is the same at all times so v = vavg at all times, therefore:
v=
or
x
t
x = vt
Object moving at constant velocity covers the same displacement in the same interval of time.
 Acceleration is the change in velocity per unit time
∆𝑣 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
=
∆𝑡
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑎=
m
SI unit:  a  = s = m/s2
s
it has direction, the same as the change of velocity
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,…
 Uniformly Accelerated Motion - motion with constant acceleration
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 =
v = u + at
example:
u = 2 m/s
a = 3 m/s
𝑣−𝑢
𝑡
→
velocity v at any time t = initial velocity u
increased by a, every second
t (s)
0
1
2
3
v (m/s)
2
5
8
11
speed increases 3 m/s
EVERY second.
← arithmetic sequence, so:
 In general: for the motion with constant acceleration:
v avg =
u+ v
2
𝑣𝑎𝑣𝑔 =
(2 + 5 + 8 + 11)𝑚/𝑠
(2 + 11)𝑚/𝑠
=
= 6.5 𝑚/𝑠
4
2
4
ALL TOGETHER:
 Uniform Accelerated Motion equations
 Any Motion
𝑓𝑟𝑜𝑚 𝑑𝑒𝑓𝑖𝑡𝑖𝑜𝑛: 𝑣𝑎𝑣𝑔 =
𝑥
𝑡
→ 𝑥 = 𝑣𝑎𝑣𝑔 𝑡
 Motion with constant velocity – uniform motion
v = vavg at all times, therefore:
x = vt
𝑥 = 𝑣𝑎𝑣𝑔 𝑡
𝑎𝑙𝑤𝑎𝑦𝑠 − 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑚𝑜𝑡𝑖𝑜𝑛
𝑣 = 𝑢 + 𝑎𝑡
𝑢 + 𝑣
2
𝑎 2
𝑥 = 𝑢𝑡 + 𝑡
2
𝑣𝑎𝑣𝑔 =
𝑣 2 = 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).
Acceleration can cause: 1. speeding up 2. slowing down
3. and/or changing direction
So beware: both velocity and acceleration are vectors. Therefore
• 1. if velocity and acceleration (change in velocity) are in the same direction, speed of the body is increasing.
• 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.
Examples of changing direction only:
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).
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!!!!!!!
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1. From graph displacement/position vs. time
∆𝑥
𝑠𝑙𝑜𝑝𝑒 =
∆𝑡
∆𝑥
𝑣𝑎𝑣𝑔 =
∆𝑡
Average velocity is equal to the slope
of the straight line joining two positions on
the position-time graph.
𝑣𝑎𝑣𝑔 = 0.25 𝑚/𝑠
Let us consider average velocities for time intervals
that are getting smaller and smaller.
As the interval of time from initial position P gets
smaller and smaller and eventually is almost zero
average velocity from initial postion P to final position
turns into Instantaneous velocity at position P
equal to the slope of the tangent line at P
on the position-time graph at time tP .
𝑣𝑃 = 2 𝑚/𝑠
𝑣𝑇 = −0.52 𝑚/𝑠
Example:
Time: 0.5 s – 5.5 s
slope doesn’t change
Instantaneous velocity at all times is the same as
average velocity in that time interval
v = 25/5 = 5 m/s
Time: 5.5 s – 8.5 s
slope is zero, so v = 0
Time: 8.5 s – 12.5 s
slope is – 25/2 , so v = – 12.5 m/s
Time: 0.5 s – 8.5 s
slope of the secant line is 3.1, so
vavg = 3.1 m/s
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
25
The same as:
= = 3.1 𝑚/𝑠
𝑡𝑖𝑚𝑒
8
Time: 0.5 s – 12.5 s
slope of the secant line is zero, so
vavg = 0
The same as:
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒
=
0
12
= 0 𝑚/𝑠
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1. What is the average velocity from t = 1s to t = 6s
𝑣𝑎𝑣𝑔 =
5 − 11
= −1.2 𝑚/𝑠
6−1
2. What is the instantaneous velocity at t = 5s?
𝑣=
6.5 − 0
= 1.5 𝑚/𝑠
7.5 − 3.1
Sketch velocity vs. time graph from position vs. time graph
These two graphs for velocity
are also acceptable.
The main thing is that
velocity decreases all the
time.
2. From graph velocity vs. time
First 5 seconds:
Slope of the graph =
Acceleration =
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 5𝑠−𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 0𝑠
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 5𝑠−𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 0𝑠
𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
= 10 m /s2
= 10 m /s2
Conclusion:
slope of the velocity–time graph is acceleration
Acceleration in time interval between 5s and 7s
is equal to the slope = 0
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Time interval from 5s to 7s
Area = (velocity between 5s and 7s)(time interval) = 100 m
Displacement= (velocity between 5s and 7s)(time interval) = 100 m
Conclusion:
Area under a velocity-time graph is the displacement.
During first 5 s the object has travelled: ½ x 50 x 5 = 125m
Sketch acceleration vs. time graph from velocity vs. time graph
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3. From graph acceleration vs. time
The acceleration vs. time graph for a car starting from rest.
Calculate the velocity of the car and hence draw the velocity vs. time graph.
For 0 – 2 seconds:
∆𝑣 = 𝑎𝑡 = (2
𝑚
4𝑚
) (2𝑠) =
= 𝐴𝑟𝑒𝑎
2
𝑠
𝑠
For 2 – 4 seconds:
∆𝑣 = 𝑎𝑡 = (0
𝑚
) (2𝑠) = 0𝑚/𝑠 = 𝐴𝑟𝑒𝑎
𝑠2
For 4 – 6 seconds:
∆𝑣 = 𝑎𝑡 = (−2
𝑚
) (2𝑠) = −4𝑚/𝑠 = 𝐴𝑟𝑒𝑎
𝑠2
The acceleration had a negative value, which means that the
velocity is decreasing. It starts at a velocity of 4 m/s and
decreases to 0 m/s.
Area under the graph v – t is the change in velocity.
ALL WE KNOW ABOUT GRAPHS
Velocity at some time is
equal to the slope of the
tangent line at that
position on the positiontime graph.
Acceleration at some
time is equal to the slope
of the tangent line at
that position on the
velocity-time graph.
From velocity – time graph :
Area under the
velocity-time
graph between
two
times/positions is
the displacement
covered in that
time interval.
Average velocity
between two positions is
equal to the slope of the
secant line between
these two points on the
position-time graph.
Average acceleration
between two positions is
equal to the slope of the
secant line between
these two points on the
velocity-time graph.
Area under the
acceleration-time
graph between two
times/positions is the
change in velocity
in that time interval.