M1 - Kinematics .pptx

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Preliminary Physics - Kinematics

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START
HERE

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END OF
LESSON
HERE

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Thinking About Journeys
• M1.1 Inquiry question: How is the motion of an object moving in a straight line described and predicted?
Group activity
Compare trips from Cobar to
Melbourne by various modes
of transport.

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Variation in Speed of Travel
Vehicles do not
typically travel at a
constant speed
• M1.1a describe uniform straight-line (rectilinear) motion and uniformly accelerated motion

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Speed and Direction Changes
A typical journey usually involves
changes of speed and direction

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Speed
Speed = m s-1
Distance = m
Time = s
•M1.1a describe uniform straight-line (rectilinear) motion and uniformly accelerated motion

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Speed
• Check with each other: what do the following
terms mean?
– Constant speed
– Average speed
– Instantaneous speed
– Initial speed
– Final speed
M1.1a describe uniform straight-line (rectilinear) motion and uniformly accelerated motion

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Speed
• Constant speed
– Travelling the same distance in every period of time
• Average speed
– The constant speed an object would need to travel so as to
travel the same distance in the same time
• Instantaneous speed
– Speed of an object in the instant of time we consider it.
This will vary from instant to instant.
• Initial speed
– The speed of an object when we first consider it at the
start of its journey
• Final speed
– The speed of an object at the end of its journey, when we
finish our consideration of it

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Speed
• When speed varies, at any
point in time, the
instantaneous speed of
the object can be
determined
• The speedometer on a car
shows the instantaneous
speed of the car
• The car’s odometer shows
the distance travelled
• What is the odometer
reading of this car?

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Work in pairs and answer these questions
• How fast is this car travelling?
• How long would it take to
complete a 90 km journey at
this speed?
• If this was the average speed of
the car during a year in which
the total distance travelled was
25 000 km, for what length of
time was the car being driven.
• How long would a journey to
the Moon take travelling at this
speed, given that the distance
to the Moon is 384 000 km?

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Work in pairs and answer these questions
• 45 km h-1
• s = d/t ; t = d/s ;
90 km / 45 km hr-1 = 2 hr
• 25 000 km, 45 km h-1
t = d/s ; 25 000 km / 45 km h-1
t = 555.(5) h = 556 h
• Moon 384 000 km
t = d/s ; 384 000 km / 45 km h-1
t = 8533.(3) h = 8533 h
or
8533.(3) h × 24 h / 1 day
= 355.(5) days
= 356 days

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Speed
• Worksheet on speed
– (don’t worry about velocity just yet)

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Scalar and Vector Quantities
• A vector quantity is any
quantity requiring both a
direction and a magnitude
• A scalar quantity is any
quantity requiring only a
magnitude to quantify its
measurement
• In an equation, a vector
quantity is represented by an
arrow pointing right written
above the symbol for the
quantity that is a vector, or in
printed text it may be typed in
bold print instead.
• v or v for velocity

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Brainstorm
• Scalar quantities • Vector quantities

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Symbols and Units for Scalar and Vector Quantities
Quantity Vector or Scalar Symbol for Quantity Symbol for Unit
time scalar t s
distance scalar s or r m
displacement vector s or r m (with direction)
speed (initial, final) scalar u, v ms–1
velocity vector v ms–1
(with direction)
acceleration vector, scalar a, a ms–2
(with direction)
mass scalar m kg
weight vector W N
force vector F N
energy scalar E J
kinetic energy scalar Ek J
momentum vector p kg.ms–1
impulse vector Impulse kg.ms–1
A diagram in which one or more vectors are represented by arrows is called a
vector diagram.

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Distance and Displacement
Distance and Displacement
Distance is a scalar quantity - a quantity with no associated direction
A person walking along the path AB travels 40 metres
Displacement is the straight line distance and direction of one point from another
and if north is in the direction indicated…
A person walking along the straight path A-C…
A
C
B
40 m
30 m
50 m
Walking along the path BC results in another 30 metres being travelled
travels a distance of 50 metres
The person’s final displacement is 50 m
bearing 53° relative to north
and the total distance travelled is 70 metres

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Various ways of direction

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Velocity
T
V
R
T
V
R
T
V
R Velocityav = Total Displacement
Time taken
Time taken = Total Displacement
Velocityav
Total Displacement = Velocityav × Time taken
Velocity = m s-1 (and direction)
Displacement = m (and direction)
Time = s

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Defining Average Velocity
• Average velocity is the change in displacement per
second over a specified time interval. This definition is
represented by the equation shown here.
• Instantaneous velocity is the velocity - speed and direction of
travel at a specific time
• Instantaneous and average speeds do not require any
statement of the direction being travelled
𝑣𝑎𝑣 =
Δ𝑠
Δ𝑡

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Velocity
As with speed, a car
would not travel at
the same velocity all
the time. However,
direction of travel
will change, and,
because velocity
includes direction,
even if the speed is
unchanged, the car is
is said to be
accelerating.

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Velocity
• Example
– a car, travelling at a constant velocity, travels 250
metres south in 20 seconds. Calculate its average
velocity.
• Solution
Data: Calculation:
Displacement = 250 m S Velocityav = displacement
Time = 20 s time taken
Velocityav = ? m s-1 250 m South / 20 sec
= 12.5 m s-1 South

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Velocity
• Complete velocity questions now

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Acceleration
= change in velocity
Speed Direction Both
𝑎 =
𝑣 − 𝑢
𝑡 Where:
a = acceleration in m s-2
u = initial velocity in m s-1
v = final velocity in m s-1
t = time taken for change in seconds
𝑣 = 𝑢 + 𝑎𝑡
or re-written as

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Acceleration Example
• Example
– Consider a car moving at 10 m s-1 east and which
accelerates at 50 m s-1 east over a period of 5s.
• Solution
a = change in velocity / time for change to occur
a = (final velocity – initial velocity)/time taken
a = (50 -10)/5 = 8 ms-1

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Graphs to summarise motion
• Position vs Time
– Slope = velocity
• Velocity vs Time
– Slope = acceleration
– Area = displacement
• Acceleration vs Time
– No acceleration = line stays on x-axis
– Constant acceleration = line horizontal
above/below axis
– Area = velocity
https://www.youtube.com/watch?v=7GJ
_SYM8cyU
Interpreting motion graphs (7m 30s)
https://www.youtube.com/watch?v=rYb
f_-HIJNE
Motion graphs explained (7m 12s) HSC
Physics Explained
https://youtu.be/alE2kBlWNi8?t=535
Acceleration v Time (2m 11s)

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Quick review from last lesson
• What would the velocity vs time graph look like for a ball that was pushed
up a ramp and which rolled back down?

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Ticker timer experiment
• M1.1b FHI instantaneous velocity and average velocity

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• AIM
– To determine
instantaneous and
average velocities of a
trolley using a ticker
timer
1.2m

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• Setup ticker timer and
trolley
• Run over 1.2 m
• Ticker timer = 50 Hz
– 0.02 s between each dot
– 0.10 s between 6 dots
• (that is, 5 time segments)
• Graph to measure
– Average velocity
– Instantaneous velocity
1.2m

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Sample Ticker timer results

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Ping pong ball experiment
• Drop a ping pong ball from a height.
– Allow it to bounce several times
– EXTENSION: record as video to allow you to make
quantitative measurements
• Describe its motion.
• Draw the following graphs:
– Distance vs time
– Displacement vs time
– Speed vs time
– Velocity vs time
– Acceleration vs time
• M1.1d FHI graphing distance, displacement, speed, velocity and acceleration vs time
RESULTS:
Ping pong ball graphs (6m)
HSC Hub

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Displacement = positive areas – negative areas

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Graphing motion
• M1.1e use modelling and graphs to analyse rectilinear motion, including SUVAT
(6m09s)

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SUVAT equations

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SUVAT: equations of motion
(7m45s)

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SUVAT equations
• The SUVAT equations are interlinked.
• Used when acceleration is constant.
• Decide which to use depending on available
data
– 5 SUVAT symbols
– 4 SUVAT used in each equation
– 3 SUVAT variables used in questions
• e.g. given u, t, a, find v.
• don’t use equation with s, as we don’t need it

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Steps to solving SUVAT questions
• Questions are often wordy, so pull out
relevant information into a concise list
s = 50 m
u = 0 m/s
v =
a = 1.2 m/s2
t = t
50 = 0 × t + ½ × 1.2 × t2
50 = 0.6 t2
t = sqrt (50/0.6)
t = sqrt (83.3)
t = 9.13 s

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SUVAT example

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SUVAT example – free fall
• Free fall
– acceleration due to gravity
– 9.81 m/s2 downwards.

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SUVAT example – free fall
• You might be tempted
to use v = u + at BUT
– What if you calculated
velocity incorrectly?
– The safe thing is to use
the formula without v

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SUVAT derivation by graphs

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Vector Diagrams
• Vector diagrams are used to represent quantities
that involve both direction and magnitude.
M1.1c calculate relative velocity of two objects moving along the same line using vector analysis

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Relative Velocity
Relative velocity is always found by subtracting one velocity from the other.
velocity of the ambulance relative to the vintage car
= vambulance – vvintage car = 100 – 60 = 40 km/h
velocity of the vintage car relative to the ambulance
= vvintage car – vambulance = 60 – 100 = –40 km/h
Generally
vA relative to B = vA – vB

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Relative Velocity
The fighter plane (A) is travelling west.
Take west to be the positive direction. This choice is arbitrary providing care is
taken with the +/– signs to indicate direction.
The velocity of the fighter plane (A) relative to the helicopter (B) is given by
vA relative to B = vA – vB = 1160 – (–160) = 1320 km/h.
Since the sign is positive, the velocity is to the west.

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Relative Velocity
vA relative to B = vA – vB = 100 – (–60) = 160 km/h north
In this situation, if B was stationary, it is as if A is approaching at 160 km/h,
heading north
VB relative to A = vB – vA = –60 – 100 = –160 km/h north
In this situation, it is as if A was stationary, and B is approaching at 160 km/h,
heading south
Note that the magnitude of the relative velocity stays the same

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Relative Velocity – for you to do
The car A is travelling east with a velocity of 90 km h–1 and B is
travelling west with a velocity of 60 km h–1.
What is the velocity of
a. A relative to B?
b. B relative to A?
Answer
vA relative to B = vA – vB
= 90 – (60) = 30 km/h east
vB relative to A = vB – vA
= 60 – (90) = – 30 km/h (west)

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Vector subtraction (relative velocities)

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Relativity
• When we say a car is travelling at 20 m/s, we
really mean
– relative to the Earth (RTE)
– a stationary observer (whose v = 0 RTE)
• As most of our experience IRL is RTE, we never
qualify such examples as “RTE”, just assume.
– e.g., a driver of said vehicle is also travelling at 20
m/s RTE, but would be travelling at 0 m/s RTcar

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Relative to Earth
• RTE relationships can be shown
mathematically

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Another example
• Observer velocity relative to car = 0 m/s
• Below is RTE car vs speed camera:

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• Each below is RTE

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Vectors in one dimension
M1.1c calculate the relative velocity of two objects moving along the same line using vector analysis

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Vectors in one dimension
• A vector is a quantity
requiring both a magnitude
and a direction in order that
the quantity be described
fully
e.g. displacement, velocity,
acceleration and force
• A vector can be represented
by an arrow, the length of
which is proportional to the
force and the direction of
which indicates the direction
of the vector
• Vector diagrams are useful
in solving problems involving
relative motion
20 km h–1 east
10 km h–1 east
10 km h–1 south

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Adding Velocities Using Vectors

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• Vector quantities can be added or subtracted using vector
diagrams
• A person is walking at 5 m s–1 through a train in the same
direction that the train is travelling at 10 m s–1. Calculate the
velocity of the person relative to the ground.
10 m s–1 5 m s–1

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• Vector quantities can be added or subtracted using vector
diagrams
• A person is walking at 4 m s–1 through a train in the opposite
direction that the train is travelling at 10 m s–1. Calculate the
velocity of the person relative to the ground.
10 m s–1
4 m s–1
6 m s–1

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Vector components
•M1.2a analyse vectors in 1D and 2D to: perpendicular components

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“Soak a toe”
M1.2a analyse vectors in 1D and 2D to: perpendicular components

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Resolving Vectors into components
W1 = W sin θ
W2 = W cos θ

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Resolving Vectors into components

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Resolving vectors into components
Tim throws a ball to Sally. When the ball leaves
Tim’s hand, it travels at a velocity of 5 m/s at
an angle of 30deg to the horizontal. What are
the horizontal and vertical components of the
ball when it leaves tim’s hands

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Adding perpendicular Vectors
• M1.2a analyse vectors in 1D and 2D to: perpendicular components
4 m/s east
3 m/s south
5 m/s
Use Pythagoras to determine
the resultant vector magnitude
a2 + b2 = c2
42 + 32 = c2
sqrt(16 + 9) = c
c = sqrt(25)
c = 5 m/s
Direction, use ‘SOH-CAH-TOA’ to determine
resultant vector angle.
Don’t be tempted to use the value you’ve
just calculated.
tan(θ) = O/A
tan(θ) = 3/4
θ = 37º from East, or bearing 90+37 = 127º
A journey takes you 4 m East and 3m
South. What would be a faster way
there?
Join the head of one to the base of
the other.
bearing 127º

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When not to add vectors

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Analysing motion using Vectors
Distance travelled:
1 km + 5 km + 5 km = 11 km
M1.2b vector analysis as relates to distance and displacement
A bus travels the journey below. First calculate its distance travelled.
Then, use vector addition to determine its displacement. As displacement is a
vector quantity, we must use vector addition.

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Vector ADDITION
• To add vectors, we join the tail of the second
vector to the head of the previous vector.
• To solve vector additions:
– Draw vector diagram
– Use Pythagoras to calculate magnitude of velocity
– Use tan(θ) to calculate direction
M1.2b distance and displacement of objects on a horizontal plane

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Vector ADDITION:
distance and displacement
M1.2b distance and displacement of objects on a horizontal plane

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Vector SUBTRACTION
• We join the tail of the second vector to the
head of the previous vector BUT
– We draw the second vector in the opposite
direction to the direction given.
• To solve vector subtractions:
• M1.2d relative motion of two objects on a plane
(second vector reversed)

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Vectors SUBTRACTION:
speed and velocity
• M1.2e relative motion of objects in 2D in a variety of situations

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Vector SUBTRACTION

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M1.2e relative motion of objects in 2D in a variety of situations
Motion on a Plane - Situations

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Motion as Vectors: Summary
• To solve vector problems:
• For subtraction, reverse the second vector
• M1.2e relative motion of objects in 2D in a variety of situations

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Motion of objects as they change
• describe and analyse algebraically, graphically
and with vector diagrams, the ways in which
the motion of objects changes, including (but
not limited to):
– velocity
– displacement (ACSPH060, ACSPH061)
• M1.2c describe and analyse … motion of objects including velocity and displacement.
Draw a
representation
of the motion

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at 2 m/s
at 1 m/s

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not limited to):
– velocity
Draw a
representation
of the motion
Interpret the drawing –
e.g., use gradient and
area under graph

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Gradient = y2 – y1
x2 – x1
= 14 – 6
20 – 10
= 8
10
= 4
5
Gradient = rise
run

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Change in displacement
Change in time
= Velocity
* (but only if the graph is
a straight line)

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When velocity-time gradient is a straight line, your velocity changes at a constant rate,
and therefore represents a constant acceleration.

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What about “average speed” and “average velocity”?

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not limited to):
– velocity
Use algebra
e.g., Pythagoras
/ trigonometry
Use vector diagrams (the
arrows and a scale, may also
use protractor and ruler)
As per examples in
the previous slides

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• Algebraically (doesn’t have to be to scale)

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• Using vectors (these are to scale)

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• Using vectors

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DEPTH STUDY for Kinematics
• Speed racer.
• Time allotted: 5 periods.
DEPTH STUDY - Kinematics

M1 - Kinematics .pptx

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M1 - Kinematics .pptx