Download Motion Power-point

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5
10
The bus moved away from the tree
The person is comparing the position of the bus with respect to the
position of the tree
Reference (or origin) is position of the tree
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The tree moved away from the bus.
The person is comparing the position of the tree with respect to the
position of the bus.
Reference (or origin) is position of the bus.
Both the observations are correct. The difference is what is taken as the
origin.
Motion is always relative. When one says that a object is moving, he/she
is comparing the position of that object with another object.
Motion is therefore change in position of an object with respect to
another object over time.
Kinematics studies motion without delving into what caused the motion.
Actual Path (2 km)
Direct Path (1.1 km)
Q. How much distance do you have to travel to reach school?
Q. If you were to draw a straight line between your house and school,
what would be the length of that line?
Actual Path (2 km)
Direct Path (1.1 km)
Q. How much distance do you travel in one round trip to the school?
Q. After one trip how far away are you from your home?
Distance = length of the actual path taken to
go from source to destination
Displacement = length of the straight line
joining the source to the destination or in
other words the length of the shortest path
Suppose it was given that the person started by point A and walked in a
straight line for 5 km. Can you calculate the end point of his/her
journey?
No, the person could be anywhere
on the circle of 5 km radius.
A
Unless we know the direction of the
motion we cannot calculate the end
point of the journey.
Robert and Sarah both start from their house. Robert walks 2 km to the
east while Sarah walks 1 km to the west and then turns back and walks 1
km.
Distance travelled by them is the same (2 km)
Is their displacement also the same?
No – Sarah is back home and her displacement is 0 m.
This is because direction of motion is different in both cases.
You require both distance and direction to determine displacement.
C
Distance AB = 3 km due East
Distance BC = 4 km due North
What is the distance travelled by a person who
moves from A to C via B?
What is the displacement? What is the direction of
the displacement?
A
B
Distance travelled = 7 km, Displacement = 5 km from A towards C.
Distance travelled per unit time or the
displacement per unit time.
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑠𝑝𝑒𝑒𝑑 =
𝑚𝑒𝑡𝑟𝑒𝑠/𝑠𝑒𝑐𝑜𝑛𝑑
𝑡𝑖𝑚𝑒
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑚𝑒𝑡𝑒𝑟/𝑠𝑒𝑐𝑜𝑛𝑑
𝑡𝑖𝑚𝑒
When an object is travelling along a straight
line its velocity is equal to its speed.

1. Kinematics

Kinematics Worksheet
(Notes from booklet)

HOMEWORK

Text: Chapter 1 Summary

Written up in logbook
Position – time graph using a data
logger

Newton’s 3 Laws Notes
1643 - 1727
Week 1: Worksheet 1
Due Date:

How do theme park rides and elevators
change how we feel?
Elevator Notes from
Chapman
Mass & Weight Notes



Objects that fall to the Earth all experience an
acceleration.
The acceleration due to gravity is g = 9.8
m/s2. In our tests and examinations this will
be 10.
This acceleration must be due to a force.

The acceleration of a
falling mass m is -g.


 
ag
Kinematic view
The force on the mass is
found from F = ma (action).
This gravitational force is
F = -mg.


Fgrav  mg
Dynamic view

We measure weight with a
scale that measures normal
force.
◦ W = mg

Weight is related to mass by
the gravitational field g.


A vertical acceleration can
change the weight.
Mass is unchanged.
Newton’s law of
acceleration
 F = ma
 net force, F = -mg + FN.
The normal force on the
floor is our sense of weight.
◦ Downward acceleration
reduces weight
◦ Upward acceleration
increases weight



Solve for the normal force
 -mg + FN = ma
 FN = ma + mg
 FN = m (a + g)

Apparent mass based on g
 mapp = FN / g





An elevator is accelerating
downward at 2.0 m/s2.
The person has a mass of
70 kg.
What mass is on the scale?
Add all the forces, but the
net force is – ma = FN – mg.
Solve for FN = m (g – a)
Convert to mass mapp = FN
/g
The scale shows 56 kg.

If the elevator accelerated downward at g, the
normal force would become 0.
◦ FN = m (g – a) = m (g – g) = 0


The person would feel weightless.
An object in free fall is weightless, but not
massless.

Projectile Motion Simulation
A projectile is an
object moving in two
dimensions under the
influence of Earth's
gravity; its path is a
parabola.
It can be
understood by
analyzing the
horizontal and
vertical motions
separately.
The speed in the x-direction is
constant; in the y-direction the
object moves with constant
acceleration g.
This photograph shows two balls
that start to fall at the same time.
The one on the right has an initial
speed in the x-direction. It can be
seen that vertical positions of the
two balls are identical at identical
times, while the horizontal position
of the yellow ball increases linearly.
Projectile Motion Notes
r = 100 m
The adjoining figure shows a Formula
1 racing track. A driver is did 10 laps,
what is the distance travelled by the
driver at the end of the race?
What is the displacement?
If the driver took 125.6 seconds to
complete the laps, what is his speed
and velocity in km/hr?
Distance = 2 ∗ Π ∗ 𝑟 ∗ 𝑙𝑎𝑝𝑠 = 6280 m, Displacement = 0 m
𝑑𝑖𝑠𝑎𝑛𝑐𝑒
6280
Speed = 𝑡𝑖𝑚𝑒 𝑚/𝑠 = 125.6 𝑚/𝑠 = 180 km/hr
25
A distance – time graph represents
the distance travelled with respect
to time.
Distance (m)
20
When an object covers equal
distance in every time interval, it is
said to be having uniform motion.
15
10
In an uniform motion, the speed of
the object remains constant.
5
0
0
10
30
20
Time (s)
Distance – Time graph
40
50
A stationary body is also an
example of uniform motion
1.25
20
1.0
15
0.75
Speed (m)/s
Distance (m)
25
10
0.25
5
0
0.5
0
10
30
20
Time (s)
40
50
Distance – Time graph
Distance travelled = 20 m
0
0
10
30
20
Time (s)
40
50
Speed – Time graph
Area of shaded region = 0.5 * 40 =
1.25
1.0
1.0
Velocity (m)/s
Velocity (m)/s
1.25
0.75
0.5
0.25
0
0
0.75
0.5
0.25
10
30
20
Time (s)
Velocity – Time graph
Uniform Motion
Acceleration = 0 m/s2
40
50
0
0
10
30
20
Time (s)
40
Velocity – Time graph
Non-uniform Motion
Acceleration = 0.125 m/s2
50
Rate of change of velocity
acceleration =
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
meter/second2
A body is said to be accelerating if there is a
change in velocity.
Velocity has magnitude and direction. A body
has acceleration when either of them changes.
1.25
1.0
1.0
Velocity (m)/s
Velocity (m)/s
1.25
0.75
0.5
0.25
0
0
0.75
0.5
0.25
10
30
20
Time (s)
40
Velocity – Time graph
Uniform Acceleration
Acceleration = 0.125 m/s2
50
0
0
10
30
20
Time (s)
40
Velocity – Time graph
Non-uniform Acceleration
50
25
A
Which object has the
maximum acceleration?
Velocity (m/s)
20
B
15
Which object has no
acceleration?
C
How much distance is covered
by object D in 20 seconds?
10
Explain the motion
represented by D. Given an
example of such a motion in
real life.
5
0
0
D
10
20
30
Time (s)
40
50
v
Velocity (m)/s
Initial velocity = u
Final velocity = v
Time = t
Acceleration = a
Displacement = s
Acceleration = Rate of change of
velocity
u
0
0
𝑎=
t
Time (s)
Velocity – Time graph
Uniform Acceleration
𝑣 −𝑢
𝑚/𝑠 2
𝑡
𝑎𝑡 = 𝑣 − 𝑢
𝒗 = 𝒖 + 𝒂𝒕 𝒎/𝒔
v
Velocity (m)/s
Initial velocity = u
Final velocity = v
Time = t
Acceleration = a
Displacement = s
u
0
0
t
Time (s)
Velocity – Time graph
Uniform Acceleration
Displacement = Area under the line
𝑠 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒 + 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒
1
𝑠 = 𝑢𝑡 +
𝑡 ∗ 𝑣 −𝑢 𝑚
2
(𝑣−𝑢)
But 𝑡 = 𝑎 or 𝑣 − 𝑢 = 𝑎𝑡
𝟏
𝒔 = 𝒖𝒕 + 𝒂𝒕𝟐 𝒎
𝟐
v
Velocity (m)/s
Initial velocity = u
Final velocity = v
Time = t
Acceleration = a
Displacement = s
u
0
0
t
Time (s)
Velocity – Time graph
Uniform Acceleration
Displacement = Area under the line
𝑠 = 𝑎𝑟𝑒𝑎 𝑡𝑟𝑎𝑝𝑒𝑧𝑖𝑢𝑚
1
𝑠=
𝑢+𝑣 ∗𝑡𝑚
2
(𝑣−𝑢)
But 𝑎 =t
1
(𝑣 − 𝑢)
𝑠=
𝑢+𝑣 ∗
𝑚
2
𝑎
∴ 2𝑎𝑠 = 𝑢 + 𝑣 ∗ 𝑢 − 𝑣
∴ 2𝑎𝑠 = 𝑢2 − 𝑣 2
𝒗𝟐 = 𝒖𝟐 + 𝟐𝒂𝒔