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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
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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
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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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KINEMATICS 1 - DISPLACEMENT & AVERAGE VELOCITY
1. After a tennis match, the players dash to the net to congratulate one another.
If they both run with a speed of 3m/s, are their velocities equal?
2. A mass initially at O moves 10 m to the right and then 2 m to the left.
What is the final displacement of the mass?
3. A mass initially at O, first moves 5 m to the right and then 12 m to the left. What is the total distance covered
by the mass and what is its displacement?
4. An object has a displacement of —5 m. It moves a distance to the right equal to 15 m and then a distance of 10 m to the left.
What is the total distance travelled and final displacement of the object?
5. A car starts out from O in a straight line and moves a distance of 20 km towards the right, and then returns to its starting
position 1 h later. What is the average speed and the average velocity for this trip?
6. A motor car travels on a circular track of radius a, as shown in the figure.
When the car has travelled from P to Q its displacement from P is
A. a√2 southwest.
B. a√2 northeast.
C. 3a/2 southwest.
D. 3a/2 northeast.
7. Juan is standing on the platform at a railway station. A train passes
through the station with speed 20 ms-1 in the direction shown measured
relative to the platform. Carmen is walking along one of the carriages of
train with a speed of 2.0 ms-1 measured relative to the carriage in the
direction shown.
Velocity is measured as positive in the direction shown on the diagram.
The velocity of Carmen relative to Juan is
A. -22 ms-1.
B. -18 ms-1.
C. +18 ms-1.
D. +22 ms-1.
the
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8. Four cars are on a straight road. The graph below shows the variation
with time t of the distance of each car from a fixed point.
a)
Rank the lines in order of initial starting position,
from most positive to most negative.
b)
Rank the lines in order of average velocity,
from most positive to most negative.
c)
Rank the lines in order of speed.
9. An object is moving along a straight line. The graph shows the object's
position from the starting point as a function of time.
a. In which segment(s) of the graph is the object motionless?
b. What is average velocity from time 0s to time 4s?
c. What was the velocity of the object at t = 4 s?
d. What is the average velocity from time 0s to time 6s?
e. What is the velocity at time t = 6s?
f. In which segments of the graph does the object have the greatest speed?
g. At which time (s) does the object reverse its direction of motion?
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10. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this week. She must get
from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway, a distance of 35.0 m, at
a speed of 3.50 m/s. The second hallway is filled with students, and she covers its 48.0-m length at an average speed of 1.20 m/s. The final
hallway is empty, and Suzette sprints its 60.0-m length at a speed of 5.00 m/s. a) Does Suzette make it to class on time or does she get
detention for being late again? b) Draw a distance vs. time graph of the situation.
11. An object is moving along a straight line in the positive x direction. The graph
shows its position from the starting point as a function of time. Various segments
the graph are identified by the letters A, B, C, and D.
of
a. Which segment(s) of the graph represent(s) a constant velocity of +1.0 m/s?
b. What was the instantaneous velocity of the object at the end of the eighth
second?
c. During which interval(s) did the object move in the negative x direction?
12. A pronghorn antelope has been observed to run with a top speed of 97 km/h. Suppose she runs 1.5 km with an average speed of 85
km/h, and then runs 0.80 km with an average speed of 67 km/h.
a. How long will it take (in seconds) the antelope to run the entire 2.3 km?
b. What is the antelope’s average speed during this time?
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13. A particle travels along a curved path between two points P and Q as shown. The
displacement of the particle does not depend on
a. the location of P.
b. the location of Q.
c. the distance traveled from P to Q.
d. the shortest distance between P and Q.
e. the direction of Q from P.
14.
a.
b.
c.
d.
e.
For which one of the following situations will the path length equal the magnitude of the displacement?
A jogger is running around a circular path.
A ball is rolling down an inclined plane.
A train travels 5 miles east; and then, it stops and travels 2 miles west.
A ball rises and falls after being thrown straight up from the earth's surface.
A ball on the end of a string is moving in a vertical circle.
15. A particle moves along the x axis from xi to xf . Of the following values of the initial and final coordinates, which results in the
displacement with the largest magnitude?
a. xi = 4m, xf = 6m
c. xi = –4m, xf = 2m
e. xi = –4m, xf = 4m
b. xi = –4m, xf = –8m
d. xi = 4m, xf = –2m
16. A particle moves along the x axis from xi to xf . Of the following values of the initial and final coordiantes, which results in a negative
displacement?
a. xi = 4m, xf = 6m
d. xi = –4m, xf = –2m
b. xi = –4m, xf = –8m
e. xi = –4m, xf = 4m
c. xi = –4m, xf = 2m
17. A particle moves along the x axis from xi to xf . Of the following values of the initial and final coordinates, which results in the
displacement with the largest magnitude?
a. xi = 4m, xf = 6m
c. xi = –4m, xf = 2m
e. xi = –4m, xf = 4m
b. xi = –4m, xf = –8m
d. xi = 4m, xf = –2m
18. A hiker travels south along a straight path for 1.5 h with an average velocity of 0.75 km/h, then travels south for 2.5 h with an average
velocity of 0.90 km/h. What is the hiker’s displacement for the total trip?
a. 1.1 km to the south
b. 2.2 km to the south
c. 3.4 km to the south
d. 6.7 km to the south
19. a. The graph shows displacement versus time.
What is the average velocity for line A?
b. The graph above shows displacement versus time.
What is the average velocity for line B?
c. The graph above shows displacement versus time.
What is the average velocity for line C?